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3 years ago

Please help asap 25 pts

Mathematics

2 answers:

Reika [66]3 years ago

8 0

A. When you solve for y, you will come out with x=0.1039, and x=-1.6039.

jasenka [17]3 years ago

6 0

Hey there again

the answer is...... A :)

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Tim bought some 25 cent and some 29 cent stamps. He paid 7.60 dollars for 28 stamps. How many of each type of stamp did he buy?

Citrus2011 [14]

Answer:

25 cent stamps = 13 and 29 cent stamps = 15

Step-by-step explanation:

x = 25 cent stamps and y = 29 cent stamps

x + y = 28......x = 28 - y

0.25x + 0.29y = 7.60

0.25(28 - y) + 0.29y = 7.60

7 - 0.25y + 0.29y = 7.60

-0.25y + 0.29y = 7.60 - 7

0.04y = 0.60

y = 0.60 / 0.04

y = 15 <===== 29 cent stamps

x + y = 28

x + 15 = 28

x = 28 - 15

x = 13 <===== 25 cent stamps

lets check it...

0.25x + 0.29y = 7.60

0.25(13) + 0.29(15) = 7.60

3.25 + 4.35 = 7.60

7.60 = 7.60 (correct..it checks out)

8 0

3 years ago

Solve for x, 5-x/3=-15

tatuchka [14]

The answer is x=50

Hope this help!

8 0

3 years ago

X + 8 = 9. Find the value of x.

algol [13]

Answer:

x = 1

Step-by-step explanation:

x + 8 = 9

=> x = 9 - 8

=> x = 1

6 0

2 years ago

Read 2 more answers

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.