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kotegsom [21]
3 years ago
12

Last year, an orange tree produced 36 oranges. This year, it produced fewer than 10. What are the possible values for how many f

ewer oranges were produced from the tree this year
Mathematics
1 answer:
weeeeeb [17]3 years ago
4 0

Answer:

d : [ 0 , 26 ]

Step-by-step explanation:

Given:-

- The oranges produced last year, x = 36

- The oranges produced this year, y < 10

Find:-

What are the possible values for how many fewer oranges were produced from the tree this year

Solution:-

- The possible values for how many fewer oranges were produced this year relative to last can be determined by solving an inequality that signifies a difference (d) between last year and this year production as follows:

                                 d = x - y

                                 d = 36 - (<10)

- To remove the inequality from Right hand side we we can shift it to left hand side by inverting the inequality sig, since last production is exactly known we have:

                                d≥36 - 10

                                d≥ 26

- So the number of oranges produced are at least 26 less than last year and lower bound would be if there is no orange produced this year. Hence, the range is:

                               d : [ 0 , 26 ]                          

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One year Roger had the lowest ERA​ (earned-run average, mean number of runs yielded per nine innings​ pitched) of any male p
baherus [9]

Answer:

(a) The <em>z</em>-score of Roger is -1.56 and the <em>z</em>-score of Amber is -1.13.

(b) Roger had a better year relative to his peers.

Step-by-step explanation:

If X follows N (<em>µ, σ</em>₂), then z=\frac{x-\mu}{\sigma}, is a standard normal variate with mean,      E (Z) = 0 and Var (Z) = 1. That is, Z follows N (0, 1).

Let <em>X</em> = ERA of male pitchers and <em>Y</em> = ERA of ERA of female pitchers.

It is provided that the mean and standard deviation of <em>X</em> are:

\mu_{X}=4.371\\\sigma_{X}=0.787

Also, the mean and standard deviation of <em>Y</em> are:

\mu_{Y}=4.363\\\sigma_{Y}=0.869

ER of Roger is 3.14 and ERA of Amber is 3.38.

(1)

Compute the <em>z</em>-score of Roger as follows:

z=\frac{x-\mu_{X}}{\sigma_{X}}=\frac{3.14-4.371}{0.787}=-1.56

Thus, the <em>z</em>-score of Roger is -1.56.

Compute the <em>z</em>-score of Amber as follows:

z=\frac{x-\mu_{Y}}{\sigma_{Y}}=\frac{3.38-4.363}{0.869}=-1.13

Thus, the <em>z</em>-score of Amber is -1.13.

(2)

Compute the probability of ERA's that are greater than Roger's ERA as follows:

P(X>3.14)=P(\frac{X-\mu}{\sigma}>\frac{3.14-4.371}{0.787})\\=P(Z>-1.56)\\=P(Z

This implies that 94% of the other male pitchers had an ERA higher than 3.14.

Compute the probability of ERA's that are greater than Amber's ERA as follows:

P(Y>3.38)=P(\frac{Y-\mu_{Y}}{\sigma_{Y}}>\frac{3.38-4.363}{0.869})\\=P(Z>-1.13)\\=P(Z

This implies that 87% of the other female pitchers had an ERA higher than 3.38.

As it is provided that the lower the ERA the better the pitcher, then it can be concluded that Roger had a better year relative to his peers.

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4 years ago
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James and Jeff are two brothers who love pizza. Between the two of them, they can eat 94 of a whole pizza. Convert this number o
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Answer: A 2 1/4

Step-by-step explanation:

9/4 is equal to 2 1/4

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3 years ago
Find the sum of the G.P,16^2+17^2+18^2+...+25^2
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Answer:

hello :

note : 1²+2²+3²+........+n² = n(n+1)(n+2)/6

Step-by-step explanation:

let : S = 1²+2²+3²+......+25²

       A =  16²+17²+18²+...+25²

      B =  1²+2²+3²+......+15²

      S = (1²+2²+3²+....+15²) + (16²+17²+18²+...+25²)

calculate : S   for : n = 25

S = 25(26)(27)/6 = 2925

calculate : B   for : n = 15

B = 15(16)(17)/6 = 680

so : S = B + A

     A  = S - B = 2925 - 680 = 2245


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