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Inga [223]
3 years ago
6

Help pleaseeee I need the answer ???

Mathematics
1 answer:
andrey2020 [161]3 years ago
3 0

the first one is the correct answer

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Question 4 of 10
Naddik [55]
The answer is D
real explanation: take 0.642 & divide it by -0.28.
7 0
2 years ago
Find the cost of this trip: Find the airfare for flying round trip from Denver to Detroit if a one-way ticket is $235.00, there
Rzqust [24]

5% of 235 is: 235*(5/100)= 235/20 = 11,75 or 5% of 470 is: 470/20 = 23,5

Total cost = 2*(235 + 11,75) = 2*246,75 = 493,5 or Total cost = 470 + 23,5 = 493,5

6 0
3 years ago
Read 2 more answers
Y=-155. Put in a table of 3 rows.
Alja [10]

Answer:

51.6666666667

Step-by-step explanation:

4 0
3 years ago
A teacher randomly chooses a two-person leadership team from a group of four qualified students. Three of the students, Sandra,
nata0808 [166]

Answer:

What is P(A), the probability that the first student is a girl? (3/4)

What is P(A), the probability that the first student is a girl? (3/4)What is P(B), the probability that the second student is a girl? (3/4)

What is P(A), the probability that the first student is a girl? (3/4)What is P(B), the probability that the second student is a girl? (3/4)What is P(A and B), the probability that the first student is a girl and the second student is a girl? (1/2)

The probability that the first student is a girl is (3/4), likewise for the 2nd 3rd and 4th it's still (3/4). The order you pick them doesn't matter.

However, once you're looking at P(A and B) then you're fixing the first position and saying if the first student is a girl what's the probability of the second student being a girl.

3 0
3 years ago
Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If​ women's weights are
Mariulka [41]

Answer:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(173.6,49.8)  

Where \mu=173.6 and \sigma=49.8

We are interested on this probability

P(141.8

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0
3 years ago
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