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2 years ago

If 30 Coins In Quarters And Dimes Are Worth $4.80, How Many Of Each Coin Are There?

Mathematics

1 answer:

GREYUIT [131]2 years ago

3 0

Answer:

There are 18 dimes and 12 quarters

Step-by-step explanation:

Remember that

1 Dime =$0.10

1 Quarter=$0.25

Let

x-----> the number of dimes

y-----> the number of quarters

we know that

x+y=30

x=30-y ------> equation A

0.10x+0.25y=4.80

Multiply by 100 both sides

10x+25y=480 ------> equation B

Substitute equation A in equation B and solve for y

10(30-y)+25y=480

300-10y+25y=480

15y=180

y=12 quarters

Find the value of x

x=30-y=30-12=18 dimes

therefore

There are 18 dimes and 12 quarters

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Which operation results in a binomial? (3y ^ 6 + 4) (9y ^ 12 - 12y ^ 6 + 16)

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Step-by-step explanation:

Given the expression, $(3y ^ 6 + 4) (9y ^ {12} - 12y ^ 6 + 16)$ , we want to know which sign placed in between the two polynomial will produce a binomial.

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$= (3y ^ 6 + 4) + (9y ^ {12} - 12y ^ 6 + 16)\\= (3y ^ 6 + 4) + (9y ^ {12} - 12y ^ 6 + 16)\\collect \ like \ terms\\= 9y^{12} +3y^6 - 12y^6 + 16+ 4\\= 9y^{12} - 9y^6 + 20\\$

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Using a plus (-) sign to check:

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Answer:

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is

$\sigma = 12.6 \: hours$

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

$n = 250$

The sample mean is

$\bar{x} = 9 \: hours$

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

$\mu = \bar{x} = 9 \: hours$

The sample standard deviation would be

$s = {\frac{\sigma}{\sqrt{n} }$

Where is the population standard deviation and n is the sample size.

$s = {\frac{12.6}{\sqrt{250} }$

$s = 0.7969 \: hours$

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

$\text {confidence interval} = \mu \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sampling size, s is the sample standard deviation and is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

$MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\$

So the required 95% confidence interval is

$\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569$