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4 years ago

Which phrase best describes the location of 22/5 on a number line?

Mathematics

1 answer:

Leona [35]4 years ago

8 0

22/5 = 4 2/5 = 4.4

Answer: 22/5 is closer to 4 than 5.

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What’s is sherry error?

beks73 [17]

Answer:

the answer is B.

Step-by-step explanation:

It makes the problem solve smoother.

3 0

3 years ago

How to estimate

monitta

4 3/5 ≈ 5
1 1/5 ≈ 1

(4 3/5) +(1 1/5)
≈ 5 +1
≈ 6 . . . . . . . . . . . the estimated sum rounded to whole numbers

_____
Additional information:
As a refinement to your estimate, you can estimate the error. The first number is 2/5 less than 5. The second number is 1/5 more than 1, so the errors add up to
.. (2/5) +(-1/5) = 1/5

That is, your "estimate" of 6 is actually too high by 1/5. You can refine that estimated value by subtracting 1/5. 5 4/5 would be a closer "estimate."

6 0

3 years ago

An exercise cube measures 27 in. on each side. What is its volume? a. 22, 785 in3 c. 18,659 in3 b. 19,683 in3 d. 17, 257 in3

Evgesh-ka [11]

Answer:

your answer is 19,683 in^3

Step-by-step explanation:

i don't know what the heck an exercise cube is, but hey, it's the world of real world problems

volume of a cube = side length ^ 3

= 27*27*27

=19,683 in^3

4 0

4 years ago

An engineer designs a new cargo ship to transport 12,000 standard shipping containers. The ship's cargo hold and a shipping cont

snow_tiger [21]

Answer:

Vol= $450,000m^3$

Step-by-step explanation:

Volume of rectangular prism is obtained using the formula:

$V=whl\\w-width\\h-height\\l-length$

Dimensions of shipping containers is given as:

$w=2.5m\\h=2.5m\\l=6m\\$

To obtain the volume of the cargo ship, we need to calculate the volume of 1 unit of a shipping container then multiply it by the number of containers the ship can carry.

let n be the number of containers ship can carry.

$V_c=whl\\V_c=2.5m\times2.5m\times6m\\V_c=37.5m^3\\$

Volume of ship, $V_s$

$V_s=nV_c$

But n=12000

$V_s=12000\times37.5m^3\\=450,000m^3$

8 0

3 years ago

Electricity usage data consists of 45 months has a mean number of units consumed is 390.47 per month with a standard deviation o

neonofarm [45]

Answer:

The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87

Step-by-step explanation:

We have the standard deviation for the sample. So we use the t-distribution 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 = 45 - 1 = 44

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.0141

The margin of error is:

M = T*s = 2.0141*170.5 = 343.4

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 390.47 - 343.40 = 47.07 units per month

The upper end of the interval is the sample mean added to M. So it is 390.47 + 343.40 = 733.87 units per month

The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87

6 0

3 years ago