15 1/2 - 13 5/4=<br> please answer i really need help :(

Help solve<br> 34 = 28 - 2/5X

NARA [144]

34 = 28 - 2/5X

Move +28 to the other side. Sign changes from +28 to -28

34-28=28-28-2/5x

34-28=-2/5x

6=-2/5x

Multiply both sides by -5/2

6(-5/2)=-2/5(-5/2)x

Cross out 6 and 2, divide by 2 and becomes :

3*-5=-15=X

Answer: X=-15

3 0

3 years ago

What is 5/9 times 32?

Ostrovityanka [42]

160/9 or in decimal form it is 17.7 and the 7 repeats

8 0

4 years ago

a shop only buys bananas with the lengths between 13 cm and 20 cm. estimate the fraction of the 80 bananas that the shop buys

ddd [48]

Answer:

44/80

Step-by-step explanation:

draw a line from 13cm upwards to the graph line and then draw a horizontal line from that point to the y axis to find that you get 18 bananas

then draw a line from 20cm upwardsto the graph line and then draw a horizontal line from that point to the y axis to find that you get 62 bananas

62-18=44 you do this because the shop buys banana lengths between 13 and 20

only from 13 to 20 = 44

so out of 80 (the total) it should be 44/80

4 0

4 years ago

Read 2 more answers

Classify the following triangle. check all that apply

Karo-lina-s [1.5K]

Answer:

Options (A), (B)

Step-by-step explanation:

Option A.

Two sides of the given triangle are equal. Therefore, the given triangle is an isosceles triangle.

Option (B).

Given triangle has a right angle. Therefore, it's a right triangle.

Option (C).

Only two sides of the given triangle are equal. Therefore, it's not an equilateral triangle.

Option (D).

Although this triangle has two acute angles (less than 90°) but the third angle is 90°. Therefore, this triangle is not an acute triangle. It's a right triangle.

Option (E).

A scalene triangle is a triangle having all sides different in measure but the given triangle is an isosceles triangle. Therefore, it's not a scalene.

Option (F).

Any angle of this triangle is not more than 90°. Therefore, the given triangle is not an obtuse triangle.

5 0

3 years ago

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

15 1/2 - 13 5/4= please answer i really need help :(