What is 78 radishes planted in a 13-foot long row written as a unit rate per inch

Answers

Answer 1

Answer: To convert to a unit rate per inch or find how many radishes per inch. You need to know how many radishes and how many inches.

There are 78 radishes.

The 13 feet converts to 156 inches (13 x 12)z

78 radishes/156 inches =
1/2 radish per inch

The unit rate is 1/2 radish per inch.

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Maria bought bought 1 2/5 pounds of lunch meat for $3.36.How much would 2 3/4 pounds of lunch meat cost?
Enter your answer in the box.
$_______

Answers

Answer:

$6.6

Step-by-step explanation:

We are given that Maria bought 1 2/5 pounds or 7/5 pounds of lunch meat for $3.36.

Now to find how much would 2 3/4 pounds or 11/4 pounds of lunch meat cost, we will have to first find how much would 1 pound of lunch meat cost.

In order to find that we will apply unitary method here.

So cost of 7/5 pounds of meat = $3.36

Cost of 1 pound of meat = 3.36/(7/5) = $2.4

Cost of 11/4 pounds of meat = 2.4 * (11/4) = $6.6

So the cost of 2 3/4 pounds of lunch meat would be $6.6.

Answer:

the answer is 6.6

Step-by-step explanation:

i took the quiz and got it right

10x² - 43x+28

How do you factor this

Answers

Keep your head up.

Trying to spread positivity

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought in to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.(a) H0: μ ≤ 5 days versus H1: μ > 5 days. At α = .05, choose the right option. Reject H0 if tcalc > 1.7960
Reject H0 if tcalc < 1.7960

b. Calculate the Test statistic.

c-1. The null hypothesis should be rejected.
i. TRUE
ii. FALSE

c-2. The average repair time is longer than 5 days.
i. TRUE
ii. FALSE

c-3 At α = .05 is the goal being met?
i. TRUE
ii. FALSE

Answers

Answer:

a) Reject H0 if tcalc > 1.7960

b) $t=(5.42-5)/((1.16)/(√(12)))=1.239$

c-1) ii. FALSE

c-2) ii.FALSE

c-3)i. TRUE

Step-by-step explanation:

1) Data given and notation

$\bar X=5.42$ represent the mean time for the sample

$s=1.16$ represent the sample standard deviation for the sample

$n=12$ sample size

$\mu_o =5$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 5 days, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

We don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Rejection zone

On this case we need a critical value that accumulates 0.05 of the area on the right tail. The degrees of freedom are given by 11. And we can use the following excel code to find the critical value : "T.INV(1-0.95,11)" and the critical value would be given by $t_(\alpha/2)=1.795$ .

And the rejection zone is given by:

Reject H0 if tcalc > 1.7960

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=(5.42-5)/((1.16)/(√(12)))=1.239$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

Since is a one side test the p value would be:

$p_v =P(t_((11))>1.239)=0.121$

c-1. The null hypothesis should be rejected.

ii. FALSE

c-2. The average repair time is longer than 5 days.

ii. FALSE

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and the true mean is not significantly higher than 5.

c-3 At α = .05 is the goal being met?

i. TRUE

We fail to reject the null hypothesis so then the goal is met.

If x1, x2, . . . , xn are independent and identically distributed random variables having uniform distributions over (0, 1), find (a) e[max(x1, . . . , xn)]; (b) e[min(x1, . . . , xn)].

Answers

Denote by $X_((n))$ the maximum order statistic, with $X_((n))=\max\{X_1,\ldots,X_n\}$ , and similarly denote by $X_((1))$ the minimum order statistic. Then the CDF for $X_((n))$ is

$F_{X_((n))}(x)=\mathbb P(X_((n))\le x)$

In order for there to be some $x$ that exceeds the value of $X_((n))$ , it must be true that $x$ exceeds the value of all the $X_i$ , so the above is equivalent to the joint probability

$F_{X_((n))}(x)=\mathbb P(X_1\le x,\ldots,X_n\le x)$

and since the $X_i$ are i.i.d., we have

$F_{X_((n))}(x)=\mathbb P(X_1\le x)\cdots\mathbb P(X_n\le x)=\mathbb P(X_1\le x)^n$
$\implies F_{X_((n))}(x)=F_X(x)^n$

where $X\sim\mathrm{Unif}(0,1)$ . We have

$F_X(x)=\begin{cases}0&\text{for }x<0\nx&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$

and so

$F_{X_((n))}(x)=\begin{cases}0&\text{for }x<0\nx^n&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$
$\implies f_{X_((n))}(x)=\begin{cases}nx^(n-1)&\text{for }0<x<1\n0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_((n))]=\displaystyle\int_0^1xnx^(n-1)\,\mathrm dx=n\int_0^1x^n\,\mathrm dx=\frac n{n+1}$

Using similar reasoning, we can find the CDF for $X_((1))$ . We have

$F_{X_((1))}(x)=\mathbb P(X_((1))\le x)=1-\mathbb P(X_((1))>x)$
$F_{X_((1))}(x)=1-\mathbb P(X_1>x,\ldots,X_n>x)=1-\mathbb P(X_1>x)^n$
$F_{X_((1))}(x)=1-(1-\mathbb P(X\le x))^n=1-(1-F_X(x))^n$
$\implies F_{X_((1))}(x)=\begin{cases}0&\text{for }x<0\n1-(1-x)^n&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$
$\implies f_{X_((1))}(x)=\begin{cases}n(1-x)^(n-1)&\text{for }0<x<1\n0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_((1))]=\displaystyle\int_0^1xn(1-x)^(n-1)\,\mathrm dx=\frac1{n+1}$

Final answer:

The expected values of the maximum and minimum of independent and identically distributed (iid) uniform random variables, x1, x2, ..., xn, are given by E[max(x1, ..., xn)] = n / (n + 1) and E[min(x1, ..., xn)] = 1 / (n + 1) respectively.

Explanation:

In mathematics, particularly in probability theory and statistics, the question is related to independent and identically distributed (iid) random variables with a uniform distribution. The expected value or mean (E) of the maximum (max) and minimum (min) of these random variables is sought.

(a) The expected value of the max of 'n' iid uniform random variables, x1, x2, ..., xn, is calculated by integrating the nth power of x from 0 to 1. It can be found via the equation E[max(x1, ..., xn)] = n / (n + 1).

(b) Similarly, the expected value of the min of 'n' iid uniform random variables is acquired by doing (1 / (n + 1)). Hence, E[min(x1, ..., xn)] = 1 / (n + 1).

By understanding these, you could visualize the various outcomes of the random variables and their distributions, demonstrating how likely each outcome could occur.

Learn more about the Expected Value of Random Variables here:

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A cube has an edge of 3 feet. The edge is increasing at the rate of 4 feet per minute. Express the volumeof the cube as a function of m, the number of minutes elapsed.
Hint: Remember that the volume of a cube is the cube (third power) of the length of a side.

Answers

The volume of the cube can be expressed as a function of m, the number of minutes elapsed as V(m) = (3 + 4m)³ feet³.

What do we mean by volume?

The volume of an object is the total space occupied by the object in the three-dimensional space.

How do we solve the given question?

In the question, we are asked to express the volume of a cube as a function of m, the number of minutes elapsed. We are given that the initial length of the edge of the cube was 3 feet, and it increases at a rate of 4 feet per minute.

∴ We can say that the length of the edge after m minutes = 3 + 4m

As 4 feet is the increase per minute and m minutes have elapsed.

We know, that the volume of a cube is the cube of the length of the edge.

∴ The volume = (3 + 4m)³.

Hence, the function of the volume of the cube in m can be written as,

V(m) = (3 + 4m)³ feet³.

Learn more about volumes at

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Answer:

so it increases by 4 per minute

the volume of a cube is v=side^3

so the side length be 5+4m where m is the number of minutes

therefor the volume can be expressed as V(m)=(5+4m)³

Step-by-step explanation:

The lines x=0, y=2x-5, and y=mx+9 form a right triangle. find the two possible values of m

Answers

Answer: -1/2, 0

Step-by-step explanation:

x=0 , y=2x-5 and y=mx+9 form a right angle. So, y=mx+9 is either perpendicular to x=0 or y=2x-5 because the angle between x=0 and y=2x-5 is not 90 degrees. The product of slopes of perpendicular lines equals to -1.

Thus, m=-1/2 if it is perpendicular to y=2x-5. And y=c, for some constant if it is perpendicular to x=0. So y=0x+9 fits the bill. (m=0, in this case)

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