The perimeter of a rectangle is 38 inches, and its area is 70 square inches. Find the length and width of the rectangle.

Suppose that 10 sophomores, 3 juniors, and 10 seniors are candidates for a prestigious mathematics award of which three will be

faltersainse [42]

Answer:

1771 possible ways

Step-by-step explanation:

In this case, we need to know first how many candidates are in total:

10 + 3 + 10 = 23 candidates in total.

Now, we need to choose 3 of them to receive an award. In this case, we have several scenarios, but as it's an award we can also assume that the order in which the candidates are chosen do not matter, so, the formula to use is the following:

C = m! / n! (m - n)!

Where m is the total candidates and n, is the number of candidates to be chosen. Replacing this data we have:

C = 23! / 3! (23 - 3)!

C = 2.59x10^22 / 6(2.43x10^18)

C = 1771

So we have 1771 ways of choose the candidates.

7 0

3 years ago

If m<1=105 and m<4=115, then m<6=<br><br> A.75 B.40 C.50 D.60

RSB [31]

The answer is 40 so B

4 0

3 years ago

A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29.

zmey [24]

Answer:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

$df=n-1=40-1=39$

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

$p_v =P(t_{(39)}>2.18)=0.0177$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X=2.2$ represent the sample mean

$s=0.29$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =2.1$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 2,1, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2.1$

Alternative hypothesis: $\mu > 2.1$

If we analyze the size for the sample is > 30 but 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:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (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".

Calculate the statistic

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

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

P-value

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

$df=n-1=40-1=39$

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

$p_v =P(t_{(39)}>2.18)=0.0177$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

7 0

3 years ago

If I put a different sticker on the 30th, 50th, and 60th pages will any of the first 600 pages have all 3 stickers?

icang [17]

Answer:

so the which page that will have the 3 sticker, we must solve the least common multiple of 30, 50, 60. A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24, ....

The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.

so the least common multiple of 30, 50 and 60 is 300. so the page that will have 3 stickers is 300th page

Step-by-step explanation:

8 0

3 years ago

Madison has three jobs. This week she earned $124 waitressing, $30 delivering packages, and $46 babysitting. What percent of her

Likurg_2 [28]

Answer:

23%

Step-by-step explanation:

7 0

3 years ago

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