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3 years ago

6

Mr Day has 113 frog stickers. He wants to give 10 frog stickers to each student. How many students will get 10 stickers? will th

ere be any left over?

2 answers:

hodyreva [135]3 years ago

8 0

113 / 10 = 11 remainder 3

11 students get 10 stickers each There are 3 left over.

Kazeer [188]3 years ago

5 0

11 student will get 10 stickers and 3 will be <span>left over.</span>

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anygoal [31]

Answer:

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Step-by-step explanation:

Given

Shape: Rhombus

$JL = 48$

$KM = 20$

Required

Determine the perimeter

The given parameter are the diagonals of the rhombus.

The perimeter (from diagonals) is calculated as thus:

$P = 2\sqrt{(JL)^2 + (KM)^2}$

Substitute values for JL and KM

$P = 2\sqrt{48^2 + 20^2}$

$P = 2\sqrt{2304 +400}$

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3 years ago

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Answer:

H 0 : μ = 3.1 H 1 : μ > 3.1

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The p value is:

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The significance level is: 0.1 or 10%

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Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1 H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

$\bar X = 3.13 , s =0.03 , n =75$

The statistic to check the hypothesis is given by:

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Replacing the info given we got:

$t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66$

The degrees of freedom are given by:

$df = n-1=75-1 =74$

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$p_v = P(t_{74} >8.66) =3.65x10^{-13}$

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

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