All categories

3 years ago

5 magazines and 3 books cost $41.50. 2 magazines and 5 books cost $42.25. Find the cost of each item.

Mathematics

1 answer:

WINSTONCH [101]3 years ago

6 0

Answer:

The Magazines each costed $8.30 and the books cost $13.80 for the amount of $41.50. and the 2 magazines and 5 books for $42.25 the 2 magazines is $21.12 for each and $8.45 for each book.

Step-by-step explanation:I got the answers by taking the total cost and taking the number of books or if it was a a magazine and I would divide it by lets just say you have 42.25 dollars you would divide that by 3 and you would do it for every item with the total cost.

You might be interested in

jonathan had 72 pieces of recycling.he sorted them into 9 bins if each bin had the same amount , how many pieces of recycling we

Mkey [24]

8, 72 pieces of recycling sorted into 9 bins will make 8 in each bin.

8 0

3 years ago

Read 2 more answers

Are 2/7 and 6/21 equivalent ratios

ra1l [238]

6 / 2 = 3
21 / 7 = 3

Yes, 2/7 and 6/21 are equivalent ratios. The second number is simply the first number tripled.

5 0

3 years ago

Read 2 more answers

Easy math i'll mark brainliest

german

Answer:

Q6: B Q7: D Q8: C Q9: D Q10: A

7 0

3 years ago

Read 2 more answers

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

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".

Solution to the problem

The population proportion have the following distribution

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

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 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

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.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago

Help me please I dont understand

Vaselesa [24]

Answer:

42°

Step-by-step explanation:

This is right triangle and sum of 2 angles is 90°:

y+48°=90°

so y= 90°- 48°= 42°

4 0

3 years ago