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fgiga [73]
4 years ago
14

. Sabrina jogged 54 laps around the

Mathematics
2 answers:
ASHA 777 [7]4 years ago
8 0

Answer: 9

Step-by-step explanation:

N/A

-BARSIC- [3]4 years ago
5 0

Answer:

B) 6 times

Step-by-step explanation:

54/9=6

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Estimate the original volume of the pyramid in Fig. 15.28, given that its frustum has a 4 m by 4 m square top which is 12 m vert
puteri [66]

The volume of the pyramid is 2496 cu.m.

<h3>What is a Pyramid ?</h3>

A pyramid is a three dimensional structure which has a polygon base and  in general triangular faces which join at the top .

It is given that

A pyramid \rm frustum has a 4 m by 4 m square top

Height = 12 m

Base = 20 m

The volume of a square pyramid with a  \rm frustum is given by

V = (1/2) * ( Area of base + Area of  \rm frustum ) * height of the  \rm frustum from the base

V = 0.5 * ( 20 *20 + 4*4 ) * 12

V = 2496 cu.m

The volume of the pyramid is 2496 cu.m.

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8 0
2 years ago
A ladder, 100m long reaches a point on the high-rise building that is 80m
kow [346]

Answer:

It's 60 meters.

5 0
3 years ago
The U.S. Department of Agriculture records data on farm acreage and number of farms by county for every county in the country. T
lisabon 2012 [21]

Solution :

From the table,

Number of countries = $N_i$

Mean = $\overline Y_i$

Standard deviation = $\sigma_i$

Margin of error of B = 50,000 acres

So we determine the $\text{sample size n}$ and then allocate the four regions of $n_1, n_2, n_3$ and $n_4$.

Under the $\text{proportional allocation}$, the $\text{equation}$ for the value of $n_1$ which yields $V(\overline Y_{st}) = D = \frac{B^2}{H}$  is given by :

$n=\frac{\sum_{K=1}^L N_K \sigma^2_K}{ND+\frac{1}{N}\sum_{K=1}^L N_K \sigma^2_K}$

Let us complete the numerator

$\sum_{K=1}^L N_K \sigma^2_K = N_1\sigma_1^2 + N_2\sigma_2^2 + N_3\sigma_3^2 + N_4\sigma_4^2$

                $= 1052 \times (271)^2 +  210 \times (79)^2 + 1376 \times (244)^2 + 418 \times (837)^2 $

                = 453329920

The bound on the error of estimation is B = 50,000 acres

Hence we get

$D=\frac{B^2}{H}$

  $=\frac{50000^2}{4}$

  = 625,000,000

$n=\frac{\sum_{K=1}^L N_K \sigma^2_K}{ND+\frac{1}{N}\sum_{K=1}^L N_K \sigma^2_K}$

  $=\frac{453329920}{3056 \times 625 + \frac{1}{3056} \times 453329920}$

 = 220.24044

≈ 221 (approximately)

Therefore, the sample size under the proportional allocation from each stratum are given by :

$n_1=n\left[\frac{N_1}{\sum_{K=1}^\mu N_K}\right]$

   $=221\left[\frac{1052}{3056}\right]$

   ≈ 76

$n_2=n\left[\frac{N_2}{\sum_{K=1}^\mu N_K}\right]$

   $=221\left[\frac{210}{3056}\right]$

   ≈ 15

$n_3=n\left[\frac{N_3}{\sum_{K=1}^\mu N_K}\right]$

   $=221\left[\frac{1376}{3056}\right]$

   ≈ 100

$n_4=n\left[\frac{N_4}{\sum_{K=1}^\mu N_K}\right]$

   $=221\left[\frac{418}{3056}\right]$

   ≈ 30

Thus, we should select 76 counties from North central region, 15 counties from the north east region, 100 from south region and 30 from west region.

We can also estimate mean acreage of each county across all the county with a margin of error of 50,000 acres.

8 0
3 years ago
There is a sum of 76 and the difference is 23 what is the second number​
sattari [20]

Answer:

53

Step-by-step explanation:

6 0
3 years ago
At a school, 52 students in 6th grade were on the honor roll. If 28% of the 6th graders were on the honor roll, about how many s
Alexandra [31]

Answer:

-186

Step-by-step explanation:

4 0
3 years ago
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