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3 years ago

11

Emily is entering a bicycle race for charity. Her mother pledges $0.10 for every 0.25 mile she bikes. If Emily bikes 8 miles, ho

w much will her mother donate?

1 answer:

Arlecino [84]3 years ago

8 0

Answer:

0.10 for 1/4 of a mile so 8 would be 32 times more. Which would be .1 x 32 or $3.20

Step-by-step explanation:

Plz mark brainliest :3

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3 years ago

A perfume company claims that the mean weight of ther new perfume is at least 8.9 fluid oz. Express the null hypothesis and the

zlopas [31]

Answer:

$H0: mu = 8.9 fl oz.\\\\$

$Ha: mu ≠8.9 fluid oz$

Step-by-step explanation:

Given that A perfume company claims that the mean weight of ther new perfume is at least 8.9 fluid oz

For testing this claim, in Statistics we perform a certain measures called hypothesis testing.

For this first step is to create null and alternate hypothesis.

Normally null hypothesis would have some statistic = something

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Hence null hypothesis would be

H0: mu = 8.9 fl oz.

Alternate hypothesis would be opposite of this claim

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Ha: mu ≠8.9 fluid oz

Hence answer is

$H0: mu = 8.9 fl oz.\\\\$

$Ha: mu ≠8.9 fluid oz$

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3 years ago

The triangles below are similar. Triangle A B C. Side A C is 10 and side A B is 5. Angle C is 30 degrees. Triangle D E F. Side E

Gnoma [55]

Answer:

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$\triangle CBA {\displaystyle \sim } \triangle FED$

$\triangle BAC {\displaystyle \sim } \triangle EDF$

Step-by-step explanation:

Given

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$AC = 10$

$AB = 5$

$\angle C = 30^\circ$

$ED = 7.5$

$DF = 25$

$\angle F =30$

$\angle E =90$

Required

Which of the options is/are true:

$\triangle CBA {\displaystyle \sim } \triangle FED$ $\triangle CBA {\displaystyle \sim } \triangle FDE$ $\triangle BAC {\displaystyle \sim } \triangle EFD$

$\triangle BAC {\displaystyle \sim } \triangle EDF$ $\triangle ABC {\displaystyle \sim } \triangle DE\ F$ $\triangle ABC {\displaystyle \sim } \triangle DFE$

The given triangles, implies that:

$A {\displaystyle \sim } \ D$

$B {\displaystyle \sim } \ E$

$C {\displaystyle \sim } \ F$

By taking each sides of both triangle, one after the other; the possible similar triangles are:

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$\triangle CBA {\displaystyle \sim } \triangle FED$

$\triangle BAC {\displaystyle \sim } \triangle EDF$

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3 years ago