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sdas [7]
3 years ago
9

PLZ Give Us The Answer

Mathematics
1 answer:
Anni [7]3 years ago
7 0
The answer  is in the snip

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If there are 210 student and only 46 students get a permission slip to holiday hills what is the percent of the student who did
Eduardwww [97]
First you would find the difference of 210 and 46, so 210-46=164.
Then you will divide the solution by the total number of students, so
164 divided by 210.
164/210= about .780 or 78%
8 0
3 years ago
Jin cut a round gelatin dessert into 8 equal pieces. Seven of the pieces were eaten. What is the angle measure of the dessert th
irga5000 [103]

We have been given that Jin cut a round gelatin dessert into 8 equal pieces. Seven of the pieces were eaten. We are asked to find the angle measure of the dessert that was left.

We know that there are 360 degrees degrees in a circle all the way around the center.

Since Jin cut all pieces of same size, so angle measure of all pieces will be equal.

To find the angle measure of each piece, we will divide 360 by 8.

\text{Angle measure of each piece of dessert}=\frac{360^{\circ}}{8}

\text{Angle measure of each piece of dessert}=45^{\circ}

Therefore, the angle measure of the dessert left dessert piece is 45 degrees and option D is the correct choice.

5 0
3 years ago
Here is a graph for one of the equations in the system of two equations.
skad [1K]
I believe that it is the second and third boxes
7 0
3 years ago
Read 2 more answers
Johnny has to buy his mother a birthday present. If he spends $90 on 5 necklaces, how much was each necklace?
NARA [144]

Answer:

$450

Step-by-step explanation:

8 0
3 years ago
The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag
VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

We have to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a p-value of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{2050}{\sqrt{49}} = 574

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

5 0
3 years ago
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