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3 years ago

9

If a father 70 percent more money in his savings account than his daughter has in her savings account and the daughter has $4500

how much is in the father account?

1 answer:

timurjin [86]3 years ago

8 0

He should have $7650 in his savings account.

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Select the correct answer. Simplify. Reset Submit

DaniilM [7]

I think it is submit
Hope this helps

5 0

2 years ago

I don't understand this :( pls help

svp [43]

Hello!

So, the slope is 2, which means "up 2 over 1" because of rise over run and 2 is 2/1.

You have to find the graph that as a line passing through (-2, -3), so the answer is going to be Graph D.

Hope this helps :))

5 0

3 years ago

Read 2 more answers

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago

Stuck on question 1, can anyone help???

poizon [28]

'A' is the square root of 25. That's 5, so take A=5 with you
as you go to the next step.

B is A³. A³ means (A x A x A). We know that 'A' is 5, so 'B' is (5x 5 x 5) = 125 .
Take B=125 with you to the next step.

'C' is B - 25. We know that 'B' is 125. So C = (125 - 25) = 100 .
Take C=100 with you to the next step.

'D' is the square root of 'C'. We know that C=100, so D = √100 .
The square root of 100 is 10, so D=10.
Take D=10 with you to the next step .

'E' is D+39. We know that D=10. So E=(10+39) = 49 .
Take E=49 with you to the last step.

'F' is the square root of 'E'. We know that E=49.

7 0

3 years ago

Read 2 more answers

Factor each completely. <br><br>10a² - 9a + 2 <br><br>

Sophie [7]

Answer:

a= $\frac{2}{5}$ or $\frac{1}{2}$

Step-by-step explanation:

Equate the equation to zero

$10a^{2}$ -9a+2=0

$10a^{2}$ -5a-4a+2=0

5a(2a-1)-2(2a-1)=0

Write out the factors

(5a-2)(2a-1)=0

Then equate each factor to zero

5a-2=0 OR 2a-1=0

5a=2 OR 2a=1

a= $\frac{2}{5}$ OR a= $\frac{1}{2}$

a= $\frac{2}{5}$ or $\frac{1}{2}$

3 0

3 years ago