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2 years ago

HELP plssssssssssssssss

Mathematics

1 answer:

Allisa [31]2 years ago

4 0

Answer:

66° your welcome my friend and my friends are not going to be able to get the money

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Mr. Mckay spent $2.60 for a box of crackers and then divided them evenly between the s students in

Nookie1986 [14]

Answer:

You would go $2.60/s

then when you have your total number of students you take the total cost and divide that by the number of students that you have to figure out the cost of crackers for each student

Step-by-step explanation:

4 0

3 years ago

3. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the s

Umnica [9.8K]

Solution:

20 % profit on $ 1.20 

= $ 20/100 × 1.20

= $ 0.20 × 1.20

= $ 0.24 

Similarly, 20 % loss on $ 1.20 

= $ 20/100 × 1.20

= $ 0.20 × 1.20

= $ 0.24

Therefore, in one pipe his profit is $ 0.24 and in the other pipe his loss is $ 0.24.

Since both profit and loss amount is same so, it’s broke even.

Answer: (a)

7 0

3 years ago

I will give the BRAINIEST Find the values of x and y in the diagram below

Brrunno [24]

:=4 Y=3

-Lanaya- AT your help- Happy to help!

8 0

3 years ago

Read 2 more answers

You can stuff envelopes twice as fast as your friend. Together, you can stuff 6750 envelopes in 4.5 hours. How long would it tak

Lynna [10]

If you're twice a fast then... we'll divide 6750 by 3, that equals 2250.
Equation.. y=your friend and you would be 2y cuz you're twice as fast.
y+2y=6750,
3y=6750
y=2250

Your friend stuffed 2250 in 4.5 hrs.
You (twice as fast) stuffed 4500 in 4.5 hrs.

Working alone, your friend can stuff 2250 in 4.5hrs so it would take 13.5hrs to stuff 6750.
Equation.. (6750/2250) x 4.5hrs = 13.5hrs

You can 4500 in 4.5hrs so it would take you 6.75 hrs to stuff 6750.
Equation.. (6750/4500) x 4.5hrs = 6.75 hrs.

4 0

3 years ago

Read 2 more answers

A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of

GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.989.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

$M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23$

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

8 0

3 years ago