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bagirrra123 [75]
3 years ago
8

How far away was the naturalist when he first saw the elephant

Mathematics
1 answer:
zlopas [31]3 years ago
4 0

Answer:

10.11 feet

Step-by-step explanation:

i know because i did this before

can i have the crown???

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(7y - 2n) - (5y + 3a) =<br><br> Pls help
Sati [7]

Answer:

2y-2n-3a

Step-by-step explanation:

(7y-2n) - (5y+3a)

7y-2n-5y-3a

2y-2n-3a

4 0
3 years ago
Read 2 more answers
17. When a single set of values is randomly divided into two equal groups, explain how the means of these two groups may be very
joja [24]

If you take 2 groups of equal cardinality, it could happen that, for example, many of the higher values go to the first group and therefore, many of the low values go to the second group, making their respective means quite different and different from the original sample mean. This could even go worse due to the possible existence of outliers, that is, values that are far different than the sample mean. An outlier tend to disrup the mean of a sample, but for smaller samples the result is much dramatic.

For example, let X be {1,2,3,4,5,6,7,8,9,100}

The elements of X sum 145, hence the mean of X is 14,5. Let divide X in two groups

Y = {1,2,3,5,9}

Z = {4,6,7,8,100}

The elements of Y sum 20, so its mean is 4

The elements of Z sum 125, so its mean is 25

Both means are quite different from each other and quite different from the mean of X. Note that if we take the mean of the means the result is 4+25/2 = 14,5 which is equal to the mean of X.

7 0
3 years ago
It's a. not sure if its B. I'll give branliest!
atroni [7]

Answer:

P is complementary to x

Step-by-step explanation:

Complementary means adding to 90 degrees

P+x = 90 degrees

So P is complementary to x

4 0
3 years ago
Read 2 more answers
A box with a square base and open top must have a volume of 32,000 cm3. find the dimensions of the box that mini- mize the amoun
zmey [24]
Alrighty


squaer base so length=width, nice


v=lwh
but in this case, l=w, so replace l with w
V=w²h

and volume is 32000
32000=w²h


the amount of materials is the surface area
note that there is no top
so
SA=LW+2H(L+W)
L=W so
SA=W²+2H(2W)
SA=W²+4HW

alrighty

we gots
SA=W²+4HW and
32000=W²H

we want to minimize the square foottage
get rid of one of the variables
32000=W²H
solve for H
32000/W²=H
subsitute

SA=W²+4WH
SA=W²+4W(32000/W²)
SA=W²+128000/W

take derivitive to find the minimum
dSA/dW=2W-128000/W²
where does it equal 0?

0=2W-1280000/W²
128000/W²=2W
128000=2W³
64000=W³
40=W

so sub back
32000/W²=H
32000/(40)²=H
32000/(1600)=H
20=H

the box is 20cm height and the width and length are 40cm
7 0
3 years ago
A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5
lara [203]

Answer:

7.3% of the bearings produced will not be acceptable

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 0.499, \sigma = 0.002

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.

So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.

Larger than 0.504

1 subtracted by the pvalue of Z when X = 0.504.

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

Z = \frac{0.504 - 0.494}{0.002}

Z = 2.5

Z = 2.5 has a pvalue of 0.9938

1 - 0.9938= 0.0062

Smaller than 0.496

pvalue of Z when X = -1.5

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

Z = \frac{0.496 - 0.494}{0.002}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.0668 + 0.0062 = 0.073

7.3% of the bearings produced will not be acceptable

4 0
4 years ago
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