Solve the inequality<br> x² + 2x + 1

alexandr1967 [171]

6/10 is larger than 9/15's, good luck on your test!

6 0

3 years ago

The monthly dues for a premium membership is $15 more than the cost of a standard membership. The premium membership cost $40 pe

77julia77 [94]

40.00 for premium - 15.00 more than standard
40-15=35.00 for standard

4 0

3 years ago

A small piece of metal weighs 0.77 gram. what is the value of the digit in tenth place

Sliva [168]

The value of the digit in the tenths place is 7. The value of the digit in the hundredths place is 7. The first number to the right of the decimal is in the tenths place and the second number to the right is in the hundredths place. The 0 is in the ones place and is the first number to the left of the decimal.

Hope this helps!

6 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:

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