MATHEMATICS COLLEGE

I need help ASAP !!!

Answers

Answer 1

Answer:

Answer:

x = 7.5

Step-by-step explanation:

take 20 degree as reference angle

using cos rule

cos 20 = adjacent / hypotenuse

0.93 = 7/x

x = 7/0.93

x = 7.5

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A bag contains 4 red marbles, 1 green one, 1 lavender one, 3 yellows, and 2 orange marbles. HINT [See Example 7.] How many sets of five marbles include at most one of the yellow ones?

Answers

Answer:

266

Step-by-step explanation:

Red marbles = 4

Green marbles = 1

Lavender marbles = 1

Yellow marbles = 3

Orange marbles = 2

To pick 5 marbles with at most 1 yellow, we can pick 5 non-yellow marbles or (4 non-yellow marble and 1 yellow marble).

Picking 5 non-yellow marbles

There are 4 + 1 + 1 + 2 = 8 non-yellow marbles.

The number of ways of picking any 5 is

$n_1 = \dbinom{8}{5} = 56$

Picking 4 non-yellow marble and 1 yellow marble

The number of ways of picking any 4 non-yellow marbles is

$n_2 = \dbinom{8}{4} = 70$

The number of ways of picking any 1 yellow marble from 3 is

$n_3 = \dbinom{3}{1} = 3$

Number of ways for both = $70*3=210$

Total number of picking 5 marbles with at most 1 yellow

Therefore, total number of ways = 56 + 210 = 266

10x² - 43x+28

How do you factor this

Answers

Keep your head up.

Trying to spread positivity

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected, and their weights at birth are recorded as:
9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds.
If Alpha = 0.05,
1. What is the critical t-value?
2. What is the decision for a statistically significant change in average weights at birth at the 5% level of significance?

Answers

Answer:

1. Critical value t=±2.447

2. The null hypothesis is failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the birth weight significantly differs from 6.6 lbs.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the birth weight significantly differs from 6.6 lbs.

Then, the null and alternative hypothesis are:

$H_0: \mu=6.6\n\nH_a:\mu\neq 6.6$

The significance level is 0.05.

The sample has a size n=7.

The sample mean is M=7.56.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.18.

The estimated standard error of the mean is computed using the formula:

$s_M=(s)/(√(n))=(1.18)/(√(7))=0.446$

Then, we can calculate the t-statistic as:

$t=(M-\mu)/(s/√(n))=(7.56-6.6)/(0.446)=(0.96)/(0.446)=2.152$

The degrees of freedom for this sample size are:

$df=n-1=7-1=6$

For a two-tailed test with 5% level of significance and 6 degrees of freedom, the critical value for t is ±2.447.

As the test statistic t=2.152 is under 2.447 and over -2.447, it falls in the acceptance region, so the effect is not significant. The null hypothesis is failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the birth weight significantly differs from 6.6 lbs.

Sample mean and standard deviation calculations:

$M=(1)/(n)\sum_(i=1)^n\,x_i\n\n\nM=(1)/(7)(9+7.3+6+. . .+6.6)\n\n\nM=(52.9)/(7)\n\n\nM=7.56\n\n\ns=\sqrt{(1)/(n-1)\sum_(i=1)^n\,(x_i-M)^2}\n\n\ns=\sqrt{(1)/(6)((9-7.56)^2+(7.3-7.56)^2+(6-7.56)^2+. . . +(6.6-7.56)^2)}\n\n\ns=\sqrt{(8.32)/(6)}\n\n\ns=√(1.39)=1.18\n\n\n$

Mr. Ruiz drove 205 miles in 5 hours on Saturday and 180 miles in 4 hours on Sunday. What was his average speed, in miles per hour, for the two days

Answers

Answer:

In miles per hour was 34 mph

Step-by-step explanation:

205/5 = 41

180/4 = 27

to find the average you add both speeds then divide by how many there are.

41 + 27 = 68

68/2 = 34

Consider the experiment of rolling a dice. Find the probability of getting an even number 3) and a number that is multiple of 3.

Answers

The required probability of rolling a dice and getting an even number and a number that is a multiple of 3 is 1/6.

To find the probability of rolling a dice and getting an even number (3, 4, or 6) and a number that is a multiple of 3 (3 or 6), we need to consider the outcomes that satisfy both conditions.

There are three even numbers on a standard six-sided dice: 2, 4, and 6. Out of these three even numbers, two of them are multiples of 3 (3 and 6).

The probability of rolling an even number is 3/6 (3 even numbers out of 6 possible outcomes) or simply 1/2.

The probability of rolling a number that is a multiple of 3 is 2/6 (2 multiples of 3 out of 6 possible outcomes) or 1/3.

Now, to find the probability of both events occurring together (rolling an even number and a number that is a multiple of 3), we multiply the individual probabilities:

Probability of getting an even number AND a number that is a multiple of 3 = (1/2) * (1/3) = 1/6.

So, the probability of rolling a dice and getting an even number and a number that is a multiple of 3 is 1/6.

Learn more about probability here:

brainly.com/question/34277017

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Find the slope and the y-intercept of the line.
y=2x-7