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Pepsi [2]
3 years ago
15

claret purchased 2 handbags and a wallet the cost of each handbag was $10 more then the wallet if she spent a total of $60 for t

he 2 handbags what is the cost of the wallet
Mathematics
1 answer:
faust18 [17]3 years ago
8 0
The cost of the two bags is 25
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If F(x) = 2x^ln(x), then F'(2) =
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Answer:

eln2(x)ln(x4)/x

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Step-by-step explanation:

7 0
3 years ago
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The proportion of adult women in a certain geographical region is approximately 49.49​%. A marketing survey telephones 100100 pe
Alinara [238K]

Answer:

a) 49%

b) 49

Step-by-step explanation:

We have that:

The proportion of adult women in a certain geographical region is approximately 49%.

a) What proportion of women in the sample of 100100 would you expect to​ see?

The same as the region, that is 49%.

b) How many​ women, on​ average, would you expect to find in a sample of that​ size?

There are 100 people. In the region, the proportion of women is 49%.

So, in a sample of 100 people, you would expect to find 49 women.

8 0
3 years ago
2x+10=5y-4 solve for x
scoundrel [369]

Answer: x=2.5y-7

Step-by-step explanation:

Subtract 10 from both sides, resulting in 2x=5y-14

Divide both sides by 2 to isolate the x, resulting in x=(5/2)y-7

5 0
3 years ago
Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the pop
Andreyy89

Answer:

The smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

Step-by-step explanation:

The complete question is:

The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,103. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income. Round your answer up to the next largest whole number.

Solution:

The (1 - <em>α</em>)% confidence interval for population mean is:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

The margin of error for this interval is:

MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

The critical value of <em>z</em> for 90% confidence level is:

<em>z</em> = 1.645

Compute the required sample size as follows:

MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

      n=[\frac{z_{\alpha/2}\cdot\sigma}{MOE}]^{2}\\\\=[\frac{1.645\times 2103}{500}]^{2}\\\\=47.8707620769\\\\\approx 48

Thus, the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

3 0
3 years ago
Solve for x, if 2x-5=7
Zigmanuir [339]
2x-5=7
2x=7+5
2x=12
X=12/2
X=6

Hope this helps!
7 0
3 years ago
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