PLEASE HELPIf u can’t read it good it says: which equation matches this graph?

1. Y= -4x + 3

2. Y= 4x + 3

3. Y= -(1/4) x +3

4. Y=- (1/4)x + 3

Answers

Answer 1

Answer:

Answer:

It's might be 2 but I'm not for sure

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Answers

Answer:

Decay Problem.

Decay rate, r = 0.014
Initial Amount =120,000

$P(t)=120000(0.986)^t$
P(10)=104,220

Step-by-step explanation:

The exponential function for growth/decay is given as:

$P(t)=P_0(1 \pm r)^t, where:\nP_0$ is the Initial Population\nr is the growth/decay rate\nt is time$

In this problem:

The city's initial population is 120,000 and it decreases by 1.4% per year.

Since the population decreases, it is a Decay Problem.

Decay rate, r=1.4% =0.014
Initial Amount =120,000

Therefore, the function is:

$P(t)=120000(1 - 0.014)^t\nP(t)=120000(0.986)^t$

When t=10 years

$P(10)=120000(0.986)^10\n=104219.8\n\approx 104220 $ (to the nearest whole number)$

Calculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. (Round your answers to three decimal places.) f(x) = x9 − 9, x1 = 1.6

Answers

Answer:

Iteration 1: $x_(2)=1.446$

Iteration 2: $x_(3)=1.337$

Step-by-step explanation:

Formula for Newton's method is,

$x_(n+1)=x_n-(f\left(x_n\right))/(f'\left(x_n\right))$

Given the initial guess as $x_(1)=1.6$ , therefore value of n = 1.

Also, $f\left(x\right)=x^(9)-9$ .

Differentiating with respect to x,

$(d)/(dx)\left(f\left(x\right)\right)=(d)/(dx)\left(x^9-9\right)$

Applying difference rule of derivative,

$(d)/(dx)\left(f\left(x\right)\right)=(d)/(dx)\left(x^9\right)-(d)/(dx)\left(9\right)$

Applying power rule and constant rule of derivative,

$(d)/(dx)\left(f\left(x\right)\right)=\left(9x^(9-1)\right)-0$

$(d)/(dx)\left(f\left(x\right)\right)=9x^(8)$

Substituting the value,

$x_(1+1)=x_1-(f\left(x_1\right))/(f'\left(x_1\right))$

$x_(2)=1.6-(f\left(1.6\right))/(f'\left(1.6\right))$

Calculating the value of $f\left(1.6\right)$ and $f'\left(1.6\right)$

Calculating $f\left(1.6\right)$

$f\left(1.6\right)=\left(1.6\right)^(9)-9$

$f\left(1.6\right)=59.71947674$

Calculating $f'\left(1.6\right)$ ,

$f'\left(1.6\right)=9\left(1.6\right)^(8)$

$f'\left(1.6\right)=386.5470566$

Substituting the value,

$x_(2)=1.6-(59.71947674)/(386.5470566)$

$x_(2)=1.446$

Therefore value after second iteration is $x_(2)=1.446$

Now use $x_(2)=1.446$ as the next value to calculate second iteration. Here n = 2

Therefore,

$x_(2+1)=x_2-(f\left(x_2\right))/(f'\left(x_2\right))$

$x_(3)=1.446-(f\left(1.446\right))/(f'\left(1.446\right))$

Calculating the value of $f\left(1.446\right)$ and $f'\left(1.446\right)$

Calculating $f\left(1.446\right)$

$f\left(1.446\right)=\left(1.446\right)^(9)-9$

$f\left(1.446\right)=18.63851065$

Calculating $f'\left(1.446\right)$ ,

$f\left(1.446\right)=9\left(1.446\right)^(8)$

$f\left(1.446\right)=172.0239252$

Substituting the value,

$x_(3)=1.446-(18.63851065)/(172.0239252)$

$x_(3)=1.337$

Therefore value after second iteration is $x_(3)=1.337$

Final answer:

To calculate two iterations of Newton's Method, use the formula xn+1 = xn - f(xn)/f'(xn). Given an initial guess of x1 = 1.6 and the function f(x) = x9 - 9, calculate f(xn) and f'(xn) at x1 and then use the formula to find x2 and x3.

Explanation:

To calculate two iterations of Newton's Method, we need to use the formula:

xn+1 = xn - f(xn)/f'(xn)

Given an initial guess of x1 = 1.6 and the function f(x) = x9 - 9, we can proceed as follows:

Calculate f(xn) at x1: f(1.6) = (1.6)9 - 9 = 38.5432
Calculate f'(xn) at x1: f'(1.6) = 9(1.6)8 = 368.64
Calculate x2: x2 = 1.6 - f(1.6)/f'(1.6) = 1.6 - 38.5432/368.64 = 1.494
Repeat the process to find x3 using the updated x2 as the initial guess.

Learn more about Newton's Method here:

brainly.com/question/31910767

#SPJ3

Not sure how to graph this

Answers

Answer:

y=2x+6

Step-by-step explanation:

The slope is 2x and the y-intercept is 6. It is shown how to graph it in the attachment.

In? gambling, the chances of winning are often written in terms of odds rather than probabilities. The odds of winning is the ratio of the number of successful outcomes to the number of unsuccessful outcomes. The odds of losing is the ratio of the number of unsuccessful outcomes to the number of successful outcomes. For? example, if the number of successful outcomes is 2 and the number of unsuccessful outcomes is? 3, the odds of winning are 2:3(Note; if the odds of winning are 2/3, the propability of sucess is2/5)The odds of event occuring are 1:6. Find (a) the propability that the event will occur,(b) propability that the event will not occur.

(a)The propability that event will occur is....(TYPE AN INTEGER OR DECIMAL ROUNDED TO THE NEAREST THOUSANDTH AS NEEDED.)

(b)The propability thet the event will not occur is...(TYPE AN INTEGER OR DECIMAL ROUNDED TO THE NEAREST THOUSANDTH AS NEEDED)

Answers

Answer:

A) The probability that the event will occur $=(1)/(7)$

B)The probability that the event will not occur = $(6)/(7)$

Step-by-step explanation:

We are given that The odds of event occurring are 1:6.

So, Number of successful events = 1

Number of unsuccessful events = 6

So, Total events = 6+1=7

a)the probability that the event will occur= $\frac{\text{Favorable event}}{\text{Total event}}$

The probability that the event will occur $=(1)/(7)$

b)The probability that the event will not occur = $\frac{\text{Favorable event}}{\text{Total event}}$

The probability that the event will not occur = $(6)/(7)$

4a+2(b+5a)+7 Answer with work I would really like it if the work was written if no that is ok ANSWER ASAP

Answers

Answer:

Step-by-step explanation:

4a+2(b+5a)+7

4a+2b+10a+7

4a+10a+2b+7

14a+2b+7

The answer is 14a+2b+7

Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in order to treat peptic or digestive disorders. The distribution of plasma aluminum levels is known to be approximately normal; however its mean u and standard deviation o are not known. The mean aluminum level for the sample of n = 10 infants is found to be X = 37.20 ug/l and the sample standard deviation is s = 7.13 ug/1. Furthermore, the mean plasma aluminum level for the population of infants not receiving antacids is known to be only 4.13 ug/1.(a) Formulate the null hypothesis and complementary alternative hypothesis, for a two-sided test of whether the mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.(b) Construct a 95% confidence interval for the true mean plasma aluminum level of the population of infants receiving antacids.(c) Calculate the p-value of this sample (as best as possible), at the a=.05 significance level.(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the alternative hypothesis, at the a = .05 significance level? Interpret your conclusion: What exactly has been demonstrated, based on the empirical evidence?(e) With the knowledge that significantly elevated plasma aluminum levels are toxic to human beings, reformulate the null hypothesis and complementary alternative hypothesis, for the appropriate one-sided test of the mean plasma aluminum levels. With the same sample data as above, how does the new p-value compare with that found in part (c), and what is the resulting conclusion and interpretation?

Answers

Answer:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

b. (32.1, 42.3)

c. p-value < .00001

d. The null hypothesis is rejected at the α=0.05 significance level

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

p-value equals < .00001. The null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacidsgreatly increases the plasma aluminum levels of children.

Step-by-step explanation:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids. This may imply that being given antacids significantly changes the plasma aluminum level of infants.

b. Since the population standard deviation σ is unknown, we must use the t distribution to find 95% confidence limits for μ. For a t distribution with 10-1=9 degrees of freedom, 95% of the observations lie between -2.262 and 2.262. Therefore, replacing σ with s, a 95% confidence interval for the population mean μ is:

(X bar - 2.262\frac{s}{\sqrt{10} } , X bar + 2.262\frac{s}{\sqrt{10} })

Substituting in the values of X bar and s, the interval becomes:

(37.2 - 2.262\frac{7.13}{\sqrt{10} } , 37.2 + 2.262\frac{7.13}{\sqrt{10} })

or (32.1, 42.3)

c. To calculate p-value of the sample , we need to calculate the t-statistics which equals:

t=\frac{(Xbar-u)}{\frac{s}{\sqrt{10} } } = \frac{(37.2-4.13)}{\frac{7.13}{\sqrt{10} } } = 14.67.

Given two-sided test and degrees of freedom = 9, the p-value equals < .00001, which is less than 0.05.

d. The mean plasma aluminum level for the population of infants not receiving antacids is 4.13 ug/l - not a plausible value of mean plasma aluminum level for the population of infants receiving antacids. The 95% confidence interval for the population mean of infants receiving antacids is (32.1, 42.3) and does not cover the value 4.13. Therefore, the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly changes the plasma aluminum levels of children.

Given one-sided test and degree of freedom = 9, the p-value equals < .00001, which is less than 0.05. This result is similar to result in part (c). the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.

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