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Len [333]
3 years ago
7

A rectangular prism is packed with cubes that measure 1/4 inch on each side

Mathematics
1 answer:
AveGali [126]3 years ago
5 0
This doesn't have enough information to solve the problem, like what the length or width, or how many cubes are in the length, or something. Not enough information leads to nobody being able to solve it
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An animal shelter spends $1.50 per day to care for each bird and $8.50 per day to care for each cat. Alexander noticed that the
SashulF [63]

Answer:

The number of birds at the shelter on Tuesday is 21  .

Step-by-step explanation:

Given as :

The amount  spent for each bird care = $ 1.50

The amount spend for each cat care = $ 8.50

Total amount spend for both bird and cat care =  $ 108.00

The total number of birds and cats on Tuesday = 30

Now,

Let The number of birds = B

and The number of cats = C

According to question

B + C = 30      and           ...........1

$ 1.50 B + $ 8.50 C = $ 108                ..............2

Solve both equations

I.e 1.50 × ( B + C ) = 1.50 × 30

Or ,  1.50 B + 1.50 C = 45

So, ( 1.50 B + 8.50 C ) - (  1.50 B + 1.50 C ) = 108 - 45

Or , (1.50 B - 1.50 B ) + (8.50 C - 1.50 C ) = 63

Or, 0 + 7 C = 63

∴   C = \frac{63}{7}

I.e  C = 9

Put The value of C in eq 1

So, B = 30 - C

or, B = 30 - 9

∴   B = 21

Hence The number of birds at the shelter on Tuesday is 21  . Answer

4 0
3 years ago
Join the zoommmmmmmmmm
ddd [48]

Answer:

Ok

Step-by-step explanation:

5 0
3 years ago
Ask Your Teacher The level of nitrogen oxides (NOX) in the exhaust after 50,000 miles or fewer of driving of cars of a particula
Georgia [21]

Answer:

The level is L = 0.084

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 0.08, \sigma = 0.01, n = 36, s = \frac{0.01}{\sqrt{36}} = 0.0017

What is the level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01?

This is X when Z has a pvalue of 1-0.01 = 0.99. So it is X when Z = 2.325.

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

By the Central limit theorem

Z = \frac{X - \mu}{s}

2.325 = \frac{X - 0.08}{0.0017}

X - 0.08 = 2.325*0.0017

X = 0.084

The level is L = 0.084

4 0
4 years ago
Can u heeeelp pleeease
navik [9.2K]

Answer:

t ≤ 12

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Paul has three cube-shaped boxes. Each box is a different size
qwelly [4]

Answer:

See explanation

Step-by-step explanation:

Paul has three cube-shaped boxes. Each box is a different size and they are stacked from the largest to the smallest. Some information about the boxes is given below.

  • The combined volume of the three boxes is 1,197 cubic inches.
  • The area of one face of the medium box is 49 square inches.
  • The volume of the smallest box is 218 cubic inches less than the volume of the medium box.

1. The medium box has the area of one face of 49 square inches, then

a^2=49\\ \\a=7\ inches

is the side length.

The volume of the medium box is

a^3=7^3=343\ in^3.

2. The volume of the smallest box is 218 cubic inches less than the volume of the medium box, then the volume of the smallest box is

343-218=125\ in^3.

Ib is the side length, then

b^3=125\\ \\b=5\ inches

The area of one face is

b^2=5^2=25\ in^2.

3. The volume of the largest box is

1,197-343-125=729\ in^3,

then if c is the side length,

c^3=729\\ \\c=9\ inches.

4. The total height of the stack is the sum of all sides lengths:

a+b+c=7+5+9=21\ inches

5. Find the surface area of each box:

  • small 6b^2=6\cdot 5^2=6\cdot 25=150\ in^2;
  • medium 6a^2=6\cdot 7^2=6\cdot 49=294\ in^2;
  • large 6c^2=6\cdot 9^2=6\cdot 81=486\ in^2.

In total, Paul needs

150+294+486=930\ in^2

of wrapping paper. He has 1,000 square inches, so Paul has enough paper to wrap all 3 boxes.

8 0
4 years ago
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