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2 years ago

How do you solve B/5+9<10

Mathematics

2 answers:

belka [17]2 years ago

8 0

Answer:

inequality form: b<5

Step-by-step explanation:

neonofarm [45]2 years ago

7 0

Answer:

B<5

Step-by-step explanation:

To solve for B in the inequality,

subtract 9 from both sides of the equation.

B/5+9<10

B/5<1

multiply both sides by 5 to isolate B.

5(B/5<1)

B<5

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Find the missing length.

Mandarinka [93]

The answer is 6 because 14+6=20

Hope this helps

4 0

3 years ago

The base of a rectangular prism is a square. The height of the prism is half the length of one edge of the base. The volume of t

ki77a [65]

Answer:

The prism is 3 by 3 by 1.5.

Step-by-step explanation:

The prism has a length x and a width x. Since it is a square at the base both length and width are the same amount x. The height is "half the length of one edge of the base". Since the base is x, this makes it 1/2x.

The volume's prism is found using V = l*w*h. Substitute and simplify.

13.5 = x*x*1/2x

13.5 = 1/2 x^3

27 = x^3

3 = x

The prism is 3 by 3 by 1.5.

4 0

4 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.

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