Please help me I don’t understand

PLS HELP WITH THIS QUESTION

seropon [69]

Answer:

Around 37, first one

Step-by-step explanation:

add up all the 7th grader number, 40

125 divided by 40 is around 3.125

3.125 times 12 = 37.5

37 because there is no half a kid

3 0

2 years ago

Need help ASAP !!! And show work please

sammy [17]

Answer:

$\sqrt{40}$ or 6.32

Step-by-step explanation:

We can find the distance between two points by using this formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

where the x and y values are derived from the given points

The points given are ( -2 , 5 ) and ( 4 , 3 )

So we plug in the x and y values of those coordinates into the distance formula

$\sqrt{(4-(-2)^2+(3-5)^2} \\4-(-2)=6\\3-5=-2\\d=\sqrt{6^2+-2^2} \\6^2=36\\-2^2=4\\36+4=40\\d=\sqrt{40}$

or 6.32

6 0

2 years ago

I’ll give brainliest (worth 15 pts)

mixas84 [53]

Answer:

48

Step-by-step explanation:

( it might be wrong pls dont report me just let me kno y its wrong )

3 0

2 years ago

Ten less the quotient of a number and three

Anettt [7]

Answer:

(x/3) - 10

Step-by-step explanation:

quotient = divide

it's ten less than what the quotient is.

7 0

2 years ago

I love sharks! In fact, before I became a statistician, I wanted to be a marine biologist specializing in shark research (I even

Ad libitum [116K]

Answer:

Null hypothesis: $\mu \leq 15$

Alternative Hypothesis: $\mu >15$

We have enough evidence to reject the null hypothesis at 10% level of significance.

Step-by-step explanation:

1) Data given

n =44, representing the sample size

$\bar X=15.6ft$ represent the sample mean for the length of great white sharks

$s=2.5ft$ represent the sample standard deviation for the length of great white sharks

$\alpha =0.1$ significance level for the test

2) Formulas to use

On this case we are intereste on the sample mean for the length of great white sharks, and based on the paragraph the hypothesis are given by:

Null hypothesis: $\mu \leq 15$

Alternative Hypothesis: $\mu >15$

since we have n>30 but we don't know the population deviation $\sigma$ so we will can use the t approximation. The sample mean have the following distribution

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

Based on this the statistic to check the hypothesis would be given by:

$t=\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$

Replacing the values given we have:

$t_{calc}=\frac{15.6-15}{\frac{2.5}{\sqrt{44}}}=1.592$

We can calculate the degrees of freedom with:

$df=n-1=44-1=43$

With $\alpha$ and the degrees of freedom we can calculate the critical value, since $\alpha=0.1$ we need a value from the t distribution with 43 degrees of freedom that accumulates 0.1 of the area on the right or 0.9 of the area on the left.

We can use excel, a calculator or a table for this, calculating this value we got:

$t_{(43,critc)}=1.302$

Since our calculatesd value was $t_{calc}=1.592>t_{crit}$ , we can reject the null hypothesis at 0.1 level of significance.

Other way in order to have a criterion for reject or don't reject the null hypothesis is calculating the p value, on this case based on the alternative hypothesis the p value would be given by:

$p_v=P(t_{(43)}>1.592)=0.0594$

So then $p_v$ so we have enough evidence to reject the null hypothesis at 10% level of significance.

3 0

2 years ago