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Darina [25.2K]
3 years ago
14

Keira rented a car for 3 days. She paid $118 rent

Mathematics
1 answer:
Stells [14]3 years ago
6 0

Answer:

389

Step-by-step explanation:

3*118 + 35

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A carpenter has at most $250 to spend on lumber the inequality 8x+12y<250 represents the numbers x of 2 by 8 boards and the n
malfutka [58]

Answer:

The carpenter will not be able to buy 12  '2 by 8 boards' and 14 '4 by 4 boards'.

Step-by-step explanation:

Given:

Amount a carpenter can spend at most = $250

The inequality to represent the amount he can spend on each type of board is given as:

8x+12y

where x represents  '2 by 8 boards' and y represents '4 by 4 boards'.

To determine whether the carpenter can buy 12  '2 by 8 boards' and 14 '4 by 4 boards'.

Solution :

In order to check whether the carpenter can buy 12  '2 by 8 boards' and 14 '4 by 4 boards' ,  we need to plugin the x=12 and y=14 in the given inequality and see if it satisfies the condition or not or in other words (12,14) must be a solution for the inequality.

Plugging in  x=12 and y=14 in the given inequality

8(12)+12(14)

96+168

264

The above statement can never be true and hence the carpenter will not be able to buy 12  '2 by 8 boards' and 14 '4 by 4 boards'.

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The average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 in
Murrr4er [49]

Answer:

Probability that average height would be shorter than 63 inches = 0.30854 .

Step-by-step explanation:

We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.

Also, a random sample of 4 women from this population is taken and their average height is computed.

Let X bar = Average height

The z score probability distribution for average height is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 64 inches

           \sigma = standard deviation = 4 inches

           n = sample of women = 4

So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)

P(X bar < 63) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{63-64}{\frac{4}{\sqrt{4} } } ) = P(Z < -0.5) = 1 - P(Z <= 0.5)

                                                        = 1 - 0.69146 = 0.30854

Hence, it is 30.85%  likely that average height would be shorter than 63 inches.

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4 years ago
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