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3 years ago

the total cost for two items is $19 is the difference in the cost of the two items is $5 find the cost of each item.

Mathematics

2 answers:

lora16 [44]3 years ago

5 0

Ok so you spent 19 dollars the difference between prices is 5 dollars and your answer would be 12 and 7. If you need any help let me know.

nevsk [136]3 years ago

4 0

The answer is $7 and $12

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If Mr. Gonzalez has 751 pencils, and he wants to equally share them with 6 people, how many

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What are the coordinates of the image of vertex F after a

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3 years ago

Clayton read 40% of his new book over the weekend. What fraction of the<br> book did Clayton read?

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Step-by-step explanation:

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3 years ago

Read 2 more answers

1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several ti

uysha [10]

Answer:

1) $0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

b) $ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=0.0826052$ represent the sample mean

$\mu$ population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom, given by:

$df=n-1=5-1=4$

The Confidence level is 0.95 or 95%, and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , the critical value would be using the t distribution with 4 degrees of freedom: $t_{\alpha/2}=2.776$

Now we have everything in order to replace into formula (1):

$0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

Part 2

The original margin of error is given by:

$ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

3 0

3 years ago

Jay's net earnings are $4000 per month. Jay's monthly expenses are $3000 per month, with $2000 of that being rent. According to

Solnce55 [7]

In Five month Jay should have $15,000 in his account has emergency fund

<h3>Given Data</h3>

Earnings per month = $4000

Monthly Expenses = $3000

Monthly Rent = $2000 part of monthly expenses

After deducting monthly expenses from Monthly earning we have

4000-3000 = $1000

<h3>Determining the 5-months Emergency Funds</h3>

Since we have a monthly expenses of $3000 a five month emergency fund will be

= 3000*5

= $15,000

Learn more about emergency fund calculation here:

brainly.com/question/13420184

7 0

2 years ago