All categories

3 years ago

I need some help plzzzzzzzz

Mathematics

2 answers:

77julia77 [94]3 years ago

7 0

Answer:

where is your question what do you need help on

Step-by-step explanation:

snow_lady [41]3 years ago

4 0

Answer:

help with what?

Step-by-step explanation:

You might be interested in

What is 4/2/3 x 1/3/7 ?

Tatiana [17]

Answer: 2/63

Step-by-

6 0

3 years ago

Read 2 more answers

How do I solve this?

Tju [1.3M]

Answer: A: 3x^2y^(3/2)

Step-by-step explanation:

This can be written as

(81*x^8*y^6)^(1/4)

Then multiply each exponent by (1/4):

81^(1/4)*x^(8(1/4))y^6(1/4))

81^(1/4) = 3

x^(8(1/4)) = x^2

y^6(1/4)) = y^(3/2)

The result: 3x^2y^(3/2)

7 0

2 years ago

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.

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