All categories

3 years ago

70 is 30% of what number?

Mathematics

2 answers:

ExtremeBDS [4]3 years ago

5 0

Answer: 233.33.

Step-by-step explanation: Consider marking this answer as brainliest if it helped you out.

Afina-wow [57]3 years ago

4 0

Answer:

Step-by-step explanation:

233.33

Hope this helps =)

You might be interested in

A conjecture and the paragraph proof used to prove the conjecture are shown.

amm1812

Answer:

Here is the correct answer. I just took the test.

Step-by-step explanation:

8 0

4 years ago

How do i put -8-3x=y in slope intercept form?

masha68 [24]

Slope intercept form : y = mx + b
<span>
-8-3x=y
</span>y = -3x - 8

Answer would the be : y = -3x - 8

6 0

3 years ago

True or False: If I solve for the angle measure to the nearest 1/10th of a degree, the answer is 50.90

Lerok [7]

360 degrees in the west

5 0

3 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

PLZ HELP ME I AM SO SO SO CONFUSED PL COMON YOU GOT TO HELP ME OUT

Musya8 [376]

Answer:

#1 is the first answer then #2 is the first answer as well

Step-by-step explanation:

basically click the first answer for both of them

3 0

3 years ago