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2 years ago

8

Ms. Grant’s car gets between 20 and 22 miles per gallon, inclusive. The gasoline she uses costs between $4.20 and $4.50 per gall

on, inclusive. What is the greatest amount Ms. Grant will spend on gasoline to drive her car 200 miles

1 answer:

maria [59]2 years ago

7 0

Answer:

You need to be more clearer with the money amount not like 4.20 - 4.50

Step-by-step explanation:

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1 year ago

Simplify -4(-8x-8) in distributive property

frosja888 [35]

Answer:

32x+32

Step-by-step explanation:

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3 years ago

jason built a rectangular tool shed that is 8 meters wide and has an area of 96 square meters. what is the length of jasons tool

pishuonlain [190]

Answer:

12 meters

Step-by-step explanation:

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6 0

3 years ago

The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand

Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

$P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})$

$P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})$

$P(X > 1560) = P(Z > \dfrac{110}{220})$

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean $\mu_x$ = np = 52 × 0.3085 = 16.042

The standard deviation = $\sqrt {n \time p (1-p)}$

The standard deviation = $\sqrt {52 \times 0.3085 (1-0.3085)}$

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

$Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})$

$Pr ( Y > 20) = P(Z >1 .338)$

From z tables

P(Y > 20) $\simeq$ 0.0903

7 0

3 years ago