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4 years ago

How many cans of paint are needed to cover an area of 2200 square units if one can of paint covers in area 400 square units

Mathematics

1 answer:

sergij07 [2.7K]4 years ago

5 0

Answer:

Step-by-step explanation:

You have a total area of 2,200 sq units to cover... and each can of paint covers 400 sq units. So, you'll need:

2200 sq units / 400 sq units/can = 5.5 cans

In such problems, we always round up, since you will need 5 full cans of paint, and half of a 6th can.

So, in total you will need 6 cans of paint to cover all 2,200 sq units.

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Round 496179.784414 to the nearest thousand.

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4 years ago

Carol has 4 cups of candy corn to decorate cookie. She wants to put 1 /12 of a cup of candy on each cookie. How many cookies can

exis [7]

What we need to find is how many cookies we can make that have 1/12 cups of candy using 4 cups of candy corn.

What we know:
4 cups of candy
1/12 cup per cookie

We need to divide 4 cups of candy corn into groups of 1/12.

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4 years ago

A rectangular garden has a length of (3x + 30) units and a width of (x + 20) units. The garden also consists of a walkway, which

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3 years ago

Read 2 more answers

What is the equation of a line that contains the point (4, 0) and is parallel to x + y = 2?

slega [8]

Answer:

x + y = 4 or y = -x + 4

Step-by-step explanation:

0 = -1[4] + b

4 = b

y = -x + 4

If you want it in Standard Form:

y = -x + 4

+x +x

_________

x + y = 4 >> Line in Standard Form

* Since the rate of change [slope] is -1 and that parallel lines have SIMILAR SLOPES, -1 remains the same.

I am joyous to assist you anytime.

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4 years ago

Babies born after 40 weeks gestation have a mean length of 52.2 centimeters (about 20.6 inches). Babies born one month early hav

Sveta_85 [38]

Answer:

a) $Z = -2.88$

b) $Z = -0.96$

c) 40 weeks gestation babies

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

a. Find the standardized score (z-score), relative to all U.S. births, for a baby with a birth length of 45 cm.

Here, we use $\mu = 52.2, \sigma = 2.5$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 52.2}{2.5}$

$Z = -2.88$

b. Find the standardized score of a birth length of 45 cm. for babies born one month early, using 47.4 as the mean.

Here, we use $\mu = 47.4, \sigma = 2.5$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 47.4}{2.5}$

$Z = -0.96$

c. For which group is a birth length of 45 cm more common?

For each group, the probability is 1 subtracted by the pvalue of Z.

Z = -2.88 has a lower pvalue than Z = -0.96, so for Z = -2.88 the probability 1 - pvalue of Z will be greater. This means that for 40 weeks gestation babies a birth length of 45 cm is more common.

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3 years ago