Pre algebra help ( will give brainliest ) sorry if it's hard to see

The physical plant at the main campus of a large state university recieves daily requests to replace fluorescent lightbulbs. The

faltersainse [42]

Answer:

34%

Step-by-step explanation:

Given that the distribution of daily light bulb request replacement is approximately bell shaped with ;

Mean , μ = 45 ; standard deviation, σ = 3

Using the empirical formula where ;

68% of the distribution is within 1 standard deviation from the mean ;

95% of the distribution is within 2 standard deviation from the mean

Lightbulb replacement numbering between ;

42 and 45

Number of standard deviations from the mean /

Z = (x - μ) / σ

(x - μ) / σ < Z < (x - μ) / σ

(42 - 45) / 3 = -1

This lies between - 1 standard deviation a d the mean :

Hence, the approximate percentage is : 68% / 2 = 34%

5 0

3 years ago

If you buy a car for 15500 an average on an average a new car loses 11% of its value the moment that is driven off the lot once

AfilCa [17]

Answer:

$13,795

Step-by-step explanation:

15500/x=100/11

(15500/x)*x=(100/11)*x - we multiply both sides of the equation by x

15500=9.0909090909091*x - we divide both sides of the equation by (9.0909090909091) to get x

15500/9.0909090909091=x

1705=x

x=1705

now we have:

11% of 15500=1705

8 0

3 years ago

Which is the value of this expression when p=-2 and q=-1?

saul85 [17]

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$

8 0

3 years ago

-2, -1, -1, 0, -1, 1, 2, ...

Andrei [34K]

2,3,3,4
This is just anything right here

4 0

3 years ago

Read 2 more answers

The National Center for Health Statistics (NCHS) reports that 70%70% of U.S. adults aged 6565 and over have ever received a pneu

Marianna [84]

Answer:

11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they received a pneumococcal vaccination, or they did not. The probability of an adult receiving a pneumococcal vaccination is independent of other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of U.S. adults aged 65 and over have ever received a pneumococcal vaccination.

This means that $p = 0.7$

20 adults

This means that $n = 20$

Determine the probability that exactly 12 members of the sample received a pneumococcal vaccination.

This is P(X = 12).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 12) = C_{20,12}.(0.7)^{12}.(0.3)^{8} = 0.1144$

11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.

3 0

3 years ago