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3 years ago

A car dealership has 55 cars and 11 vans what is the ratio of cars to vans

Mathematics

2 answers:

IceJOKER [234]3 years ago

8 0

55:11 or simplified is 5:1

murzikaleks [220]3 years ago

4 0

One to Five I think. Hope this helps

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Quadrilateral ABCD is rotated 270° clockwise about the origin. What are the

nirvana33 [79]

Answer:

Step-by-step explanation:

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2 years ago

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if rectangle ABCD was reflected over the y-axis, reflected over the x-axis, and rotated 180 degrees, where would point A lie.

cupoosta [38]

Answer: Point A would be at (1,-1)

3 0

2 years ago

This is my last question by the way

Leni [432]

Part A:

There are 25 blocks for the length, and the length of each is 5.5 inches. 5.5x25=137.5

There are 9 blocks for the height, and each is 2.75 inches. 2.75x9=24.75

Since we're trying to find how much longer the length is to the width, we find the difference. 137.5-24.75=112.75

So part A answer is 112.75 (or 112¾)

Part B:

We know the length is 137.5 and the height is 24.75. To find how many times longer the length is, we need to divide to see how many 24.75 can go into 137.5.

137.5÷24.75=5.55

So the answer to part B is 5.55. Given the context, the answer would probably be 5, because that's how many entire height can fit.

Sorry if it's wrong lol

8 0

3 years ago

For the school play, 40 rows of chairs are set up. There are 22 chairs on each row. How many chairs are there?

Nataliya [291]

There are 880 chairs.
40x22=880

4 0

3 years ago

A random sample of 10 shipments of stick-on labels showed the following order sizes.10,520 56,910 52,454 17,902 25,914 56,607 21

sammy [17]

Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{340119}{10} = 34011.9$

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

$S.D = \sqrt{\frac{2711418821}{9}} = 17357.09$

Confidence interval:

$\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621$

$34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)$

7 0

3 years ago