Is this equation true, false or open? 9p+8=10p+7

I need a bit help im new to this math stuff

kompoz [17]

Answer:

I think it is the 3rd one

3 0

3 years ago

Nissan wants to administer a satisfaction survey to its current customers. Using their customerâ€‹ database, the company randoml

Irina-Kira [14]

Answer: simple random sampling

Step-by-step explanation:

A simple random sampling is a sampling method in which individuals are chosen randomly from the population. In this each individual has equal chances to get selected.

If N is the population size , then for any individual , the chances for getting selected is $\dfrac{1}{N}$ .

As per given , the company randomly selects 50 customers directly from the database without applying any further steps and asks them about their level of satisfaction with the company.

Thus , the sampling technique used here is simple random sampling.

3 0

3 years ago

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$