Evaluate the expression when b= 19 and c=5.

What does 1/2(4-14y)=y+50 equal

schepotkina [342]

Step-by-step explanation:

1/2(4-14y)=y+50

1/(8 - 28y) = y + 50

8y - 400 - 28y² +

Now you can solve

3 0

3 years ago

Read 2 more answers

The sum of three times a number n and 2

mario62 [17]

It would be 6n because 2times 3 is 6 and bring down the n

8 0

3 years ago

If 9p=5q which is the correct ratio of p:q

tatyana61 [14]

Answer: LHS = RHS

Step-by-step explanation:

Heya!!

Here, we need to compare the LHS and RHS

So, here is the solution -

(9p-5q)²+ 180 pq = (9p+5q)

LHS

= (9p-5q)²+ 180 pq

= 81 p² + 25 q² + 90 pq + 180 pq

= (9p - 5q) ²

RHS = (9p - 5q) ²

Hence, proved => LHS = RHS

Hope it helps!

4 0

3 years ago

Read 2 more answers

A school director must randomly select 6 teachers to part in a training session. There are 34 teachers at school. In how many di

prohojiy [21]

Answer:

Total possible ways to select 6 teachers from 34 teacher are $^{34}C_6=1344904$ .

Step-by-step explanation:

It is given that total number of teachers at a school is 34.

The school director must randomly select 6 teachers to part in a training session.

$^nC_r=\frac{n!}{r!(n-r)!}$

Where, n is total possible outcomes and r is number of selected outcomes.

Total teachers = 34

Selected teachers =6

Total number of possible ways to select 6 teachers from 34 teacher is

$^{34}C_6=\frac{34!}{6!(34-6)!}$

$^{34}C_6=\frac{34\times 33\times 32\times 31\times 29\times 28!}{6!(28)!}$

$^{34}C_6=1344904$

Therefore total possible ways to select 6 teachers from 34 teacher are $^{34}C_6=1344904$ .

7 0

3 years ago

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

8 0

3 years ago