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3 years ago

What equation can be used to find n, the number of nickels in the collection ?

Mathematics

1 answer:

Semenov [28]3 years ago

3 0

Given, Bodhi has a collection of 175 dimes and nickel. If she has n numbers of nickel. Then,

The number of dimes coin= 175-n.

The collection is worth $13.30.

1 nickel is worth five cents. So, n nickel will worth 0.05n.

Similarly 1 dime is worth 10 cents. So, 175-n dimes will worth 0.1(175-n).

Since the total worth is $13.30. So, we can set the equation as follows:

0.1(175-n)+0.05n=13.13.

Hence, the last option is the correct choice.

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Step-by-step explanation:

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Consider the following hypothesis test: H0: LaTeX: \mu_1-\mu_2=0μ 1 − μ 2 = 0 Ha: LaTeX: \mu_1-\mu_2\ne0μ 1 − μ 2 ≠ 0 The follow

Pavel [41]

Answer:

$z=\frac{104-\bar 106}{\sqrt{\frac{8.4^2}{80}+\frac{7.6^2}{70}}}}=-1.53$

Step-by-step explanation:

Data given and notation

$\bar X_{1}=104$ represent the mean for the sample 1

$\bar X_{2}=106$ represent the mean for the sample 2

$\sigma_{1} =8.4$ represent the population standard deviation for the sample 1

$\sigma_{2}=7.6$ represent the population standard deviation for the sample 2

$n_{1}=80$ sample size for the group 1

$n_{2}=70$ sample size for the group 2

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

H0: $\mu_{1}-\mu_{2} =0$

H1: $\mu_{1} -\mu_{2} \neq 0$

If we analyze the size for the samples both are greater than 30 and we know the population deviations so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:

And with this we can replace in formula (1) like this:

$z=\frac{104-\bar 106}{\sqrt{\frac{8.4^2}{80}+\frac{7.6^2}{70}}}}=-1.53$

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3 years ago

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yanalaym [24]

Answer:

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749,000 ÷ 1.07 = 700,000

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