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3 years ago

Caroline babysat for two families - the Nelsons and the Smiths. Use the info below to compare her earnings.

Mathematics

1 answer:

Ket [755]3 years ago

5 0

Answer:

She earns $6.50 from the Smiths which is 25 cents more than she earns at the Nelsons.

Step-by-step explanation:

To solve this problem you must find the hourly rate she earns at the Smiths. This will allow you to compare to the hourly rate that is given for the Nelsons. To find the Smiths hourly rate divide $39 by the 6 hours she babysat there. This will have the hourly rate. 39/6=6.5. This means she earns $6.50 an hour for the Smiths. When comparing the rates it is clear that she earned 25 cents more at the Smiths.

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A- 6=9. Please help

olchik [2.2K]

Answer:

a = 15

Step-by-step explanation:

6 + 9 = 15

a = 15

15 - 6 = 9

yes

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2 years ago

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Find the diameter and circumference of a circle with the radius of 13.1 km.

Alexandra [31]

R = 13.1
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2 years ago

An equation is shown.

Anit [1.1K]

We take the equation 4x+y=5 and subtract 4x from both sides to get
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3 years ago

State if the triangles in each pair are similar. If so, state how you know they are similar and complete the similarity statemen

astraxan [27]

Answer:

angle BEC is congruent to angle FEG (vertical opposite angles)

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Step-by-step explanation:

5 0

2 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 430 green peas and 155 yellow peas. Use a 0.05

Dimas [21]

Answer:

the null and alternative hypothesis are

$H_{0}$ : p=0.24

$H_{a}$ : p≠0.24

Test statistic is 1.416.

P-Value is 0.1568

We fail to reject the null hypothesis under 0.05 significance level

At 0.05 significance we cannot say that the proportion of offspring peas will be yellow is not 0.24 (24%)

Step-by-step explanation:

Let p be the proportion of offspring peas will be yellow. Then,

$H_{0}$ : p=0.24

$H_{a}$ : p≠0.24

test statistic can be calculated as

z= $\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of yellow offspring peas ( $\frac{155}{155+430} =0.265$
p is the proportion of yellow offspring peas assumed under null hypothesis. (0.24)

N is the sample size (585)

Test statistic is z= $\frac{0.265-0.24}{\sqrt{\frac{0.24*0.76}{585} } }$ ≈ 1.416.

Using z-table, we can find that P-Value is 0.1568

We fail to reject the null hypothesis under 0.05 significance level since 0.157>0.05

We can conclude that at 0.05 significance we cannot say that the proportion of offspring peas will be yellow is not 0.24 (24%)

6 0

2 years ago