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3 years ago

Please help anybody due in LESS THAN 1 MIN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

maxonik [38]3 years ago

8 0

Answer:

distance was 35 units and the average speed was 5 units per minute

Step-by-step explanation:

-47--12=-35

the absolute value of -35 is 35

35 units took 7 minutes

so 1 unit takes 35/7 minutes= 5 minutes

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3 years ago

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HELP! What does X equal? WILL GIVE BRAINLIEST!

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X is 31
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3 years ago

An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the

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Answer:

a. Z-test of a population mean.

Step-by-step explanation:

The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75.

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3 years ago

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2 years ago

A major credit card company has determined that customers charge between $100 and $1100 per month. If the monthly amount charged

Alexxx [7]

Answer:

a) $E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

b) First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

c) $P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

d) $P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

Step-by-step explanation:

For this case we define the random variable X who represent the customers charge, and the distribution for X on this case is:

$X \sim Unif (a= 100, b=1100)$

Part a

For this case the average is given by the expected value and we can use the following formula:

$E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

Part b

First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

Part c

For this case we want to find the percent between 600 and 889, so we can use the cumulative distribution function given by:

$F(x) = \frac{X -100}{1100-100}= \frac{x-100}{1000}, 100 \leq X \leq 1100$

And we can find this probability:

$P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

Part d

$P(311 < X< 889) = P(X$

And using the cdf we got:

$P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

6 0

3 years ago