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3 years ago

Is this correct ? check plsss...

2 answers:

7 0

This looks one hundred percent correct to me and to check your answer you can plug in the answer you find to the first equation

17=37+4f
17=37+4(-5)
17=37-20
17=17

great job!!

cluponka [151]3 years ago

7 0

Looks great to me! :D

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How do u do these 4 I need to make them a decimal

Tasya [4]

#5 -.66
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8 0

3 years ago

Evaluate the expression 2n - 6 if n = 16.

V125BC [204]

Answer:

$26$

Step-by-step explanation:

$2n-6$

Given that:

$n=16$

To evaluate the expression, we must simply substitute the given value of $n$ , which is $16$ , for $n$ :

$2(16)-6$

Multiply:

$32-6$

Subtract:

$26$

To check your work, simply set the given expression equal to our solution, $26$ :

$2n-6=26$

Add $6$ to both sides of the equation:

$2n=32$

Divide both sides of the equation by the coefficient of $n$ , which is $2$ :

$n=16$

Since it matches the given value of $n$ , our solution is correct!

3 0

2 years ago

Read 2 more answers

What is the difference between log(x/y) and log(x)/log(y)?

Yuliya22 [10]

They are different equations lol. If you plug in the same x and y for both you wont get the same answer. Do you have answer choices?

6 0

3 years ago

At the beginning of football season, Coach Carnes takes inventory of the team equipment to see what he needs. He counts 24 footb

Leviafan [203]

Answer:

Step-by-step explanation:

To solve this, lets first find the number of footballs he still needs:

75-24 = 51

Now to find the number of packages, divide by 4, since there are 4 balls per package:

51/4 = 12 Remainder 3

Since there is a remainder, and the Coach MUST have at least 75, then we need to add a package to include the remainder:

12 + 1 = 13 Packages

6 0

3 years ago

f the dean wanted to estimate the proportion of all students receiving financial aid to within 3% with 99% reliability, how many

Oksanka [162]

Answer:

n=1849

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: