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3 years ago

Write an equation please

Mathematics

1 answer:

Illusion [34]3 years ago

6 0

Let d represent the amount of money Donavon brought to the fair.

After spending half his money on rides, he had half of it left: d/2.

After spending $3 on food, he had left (d/2 -3).

Donavon and his 3 friends together had 4 times this amount. They had $8.

... 4(d/2 -3) = 8

_____

The solution can be found by simplifying the equation, adding the opposite of the constant on the left, then dividing by the coefficient of d.

... 2d -12 = 8 . . . . . eliminate parentheses

... 2d = 20 . . . . . . . add 12

... d = 10 . . . . . . . . divide by 2

Donavon and each of his friends brought $10 to the fair.

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4 years ago

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3 years ago

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3 years ago

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3 years ago

An analysis of sales records for the last 120 weeks gives the following results. Assuming that these past data are a reliable gu

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Answer:

(a) 0.5333

(b) 0.6583

(d) 0.7083

(e) 0.6167

Step-by-step explanation:

Denote the events as follows:

A = a competitor will advertise

NA = a competitor will not advertise

L = Low sales will be achieved

M = Medium sales will be achieved

H = High sales will be achieved

The data provided is of the form:

Low (L) Medium (M) High (H) Total

A 32 14 18 64

NA 21 12 23 56

Total 53 26 41 120

The probability of an event E is:

$P(E)=\frac{n(E)}{N}$

n (E) = favorable outcomes of event E

N = Total number of outcomes

(a)

Compute the probability that next week the competitor will advertise as follows:

$P(A)=\frac{n(A)}{N}=\frac{64}{120}=0.5333$

Thus, the probability that next week the competitor will advertise is 0.5333.

(b)

Compute the probability that next week sales will not be high as follows:

$P(H^{c})=1-P(H)=1-\frac{n(H)}{N}=1-\frac{41}{120}=\frac{120-41}{120}=0.6583$

Thus, the probability that next week sales will not be high is 0.6583.

(c)

The events of achieving a medium or high sales are mutually exclusive.

Since the sales achieved will either be medium or high. They cannot be both.

So, P (M ∩ H) = 0.

Compute the probability that next week there will be medium or high sales will be achieved as follows:

$P(M\cup H)=P(M)+P(H)=\frac{26}{120}+\frac{41}{120}=\frac{26+41}{120}=0.5583$

Thus, the probability that next week there will be medium or high sales will be achieved is 0.5583.

(d)

Compute the probability that next week either the competitor will advertise, or only low sales will be achieved as follows:

$P(A\cup L)=P(A)+P(L)-P(A\cap L)=\frac{64}{120}+\frac{53}{120}-\frac{32}{120}=\frac{85}{120}=0.7083$

Thus, the the probability that next week either the competitor will advertise, or only low sales will be achieved is 0.7083.

(e)

Compute the probability that next week either the competitor will not advertise, or high sales will be achieved as follows:

$P(NA\cup H)=P(NA)+P(H)-P(NA\cap H)=\frac{56}{120}+\frac{41}{120}-\frac{23}{120}=\frac{74}{120}=0.6167$

Thus, the the probability that next week either the competitor will not advertise, or high sales will be achieved is 0.6167.

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4 years ago