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nydimaria [60]
3 years ago
5

There were 75 roasted turkeys at Ally's Eats. Bob's Goods had 4/5 as many roasted turkeys as Ally's Eats. If Ally's Eats had 5/6

as many roasted turkeys as Charlotte's Market what was the average number of roasted turkeys owned by each of the three shops?
Mathematics
1 answer:
Vilka [71]3 years ago
8 0

Answer:

Step-by-step explanation:

Roasted turkey in Ally's Eats = 75

Roasted turkey in Bob's goods  = 4 / 5 x  Ally's Eats

= (4 / 5) x 75

= 60

Roasted turkey in Charlotte's Market = N ( let )

Given

5/6  N  = 75

N = 6 / 5 x 75

= 90

Roasted turkey in Charlotte's Market  = 90 .

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Fittoniya [83]

Answer:

\displaystyle 702

Step-by-step explanation:

we are given a exponential function

\displaystyle f(x) =  {3}^{x + 1}  - 22

where x represents the number and f(x) represents the amount

we are also given that when x is 3 then f(x) is 59 likewise when x is 6 then f(x) is 2165

to figure out the average rate of change between 3 and 6 we can consider the average rate of change formula given by

\displaystyle m =  \frac{ f(x)_ {2} -  f(x)_{1} }{ x_{2} -  x_{1} }

substitute what we have:

\displaystyle m =  \frac{ 2165 -  59 }{6 - 3 }

simplify substitution:

\displaystyle m =  \frac{2106 }{3 }

simplify division:

\displaystyle m =  702

hence, the average rate of change between 3 and 6 is <u>7</u><u>0</u><u>2</u>

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MR MARK MARKS HIS CLASS ON A NORMAL CURVE. THOSE WITH z-SCORES ABOVE 1.8 WILL RECEIVE AN A, THOSE
lukranit [14]

Using the normal distribution, it is found that the percentages are given as follows:

  • 3.59% of the grades will be A.
  • 9.98% of the grades will be B.
  • 74.92% of the grades will be C.
  • 8.64% of the grades will be D.
  • 2.87% of the grades will be F.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The prorportion of students who receive an A is <u>one subtracted by the p-value of Z = 1.8</u>.

Looking at the z-table, Z = 1.8 has a p-value of 0.9641.

1 - 0.9641 = 0.0359.

3.59% of the grades will be A.

For B, it is the <u>p-value of Z = 1.8 subtracted by the p-value of Z = 1.1</u>, hence:

0.9641 - 0.8643 = 0.0998.

9.98% of the grades will be B.

For C, it is the <u>p-value of Z = 1.1 subtracted by the p-value of Z = -1.2</u>, hence:

0.8643 - 0.1151 = 0.7492.

74.92% of the grades will be C.

For D, it is the <u>p-value of Z = -1.2 subtracted by the p-value of Z = -1.9</u>, hence:

0.1151 - 0.0287 = 0.0864.

8.64% of the grades will be D.

For F, it is the <u>p-value of Z = -1.9</u>, hence 2.87% of the grades will be F.

More can be learned about the normal distribution at brainly.com/question/15181104

#SPJ1

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