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4 years ago

What is 12 5/12 - 1 7/12?

Mathematics

1 answer:

pav-90 [236]4 years ago

6 0

I just did it in my calculator and got 10.8333333333

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PLEASE HELP I WILL MARK BRAINLIEST!

MArishka [77]

Answer: f(0)= -4, f(5)= 15, f(-3)= -7

Step-by-step explanation:

0 < 1, so plug into x-4. 0-4=-4

5> 1, so plug into 3(x). 3*5=15

-3<1, so plug into x-4. -3-4=-3+(-4)= -7

7 0

3 years ago

A line has a slope of zero and passes through the point (-2, 3). Which of the following points must also lie on the line? (-2, 5

igomit [66]

Answer:

The answer is B ---> (5,3)

Step-by-step explanation:

If a line has a slope of zero, then it is just a flat horizontal line. So, a point a that line will be the one with the same y-coordinate. Option B is the only option with a y-coordinate that matches the point in the question.

Have a good day

7 0

3 years ago

Read 2 more answers

Y=5+3x for x <br> Show work

Levart [38]

Answer:

$x = \frac{y - 5}{3}$

Step-by-step explanation:

Hello!

To solve for x, we have to first isolate the variable.

<h3>Solve for x</h3>

$y = 5 + 3x$
$y - 5 = 3x$
$y - 5 = 3(x)$
$\frac{y - 5}{3} = x$

The value of x is $\frac{y - 5}{3}$ .

5 0

2 years ago

How do I find the volume of a cylinder?

Keith_Richards [23]

Here you go my [email protected]

5 0

3 years ago

Read 2 more answers

For a certain population of sea turtles, 18 percent are longer than 6.5 feet. A random sample of 90 sea turtles will be selected

nika2105 [10]

Answer:

The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405

Step-by-step explanation:

Central Limit Theorem:

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

18 percent are longer than 6.5 feet.

This means that $p = 0.18$

A random sample of 90 sea turtles

This means that $n = 90$

What is the standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.18*0.82}{90}} = 0.0405$

The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405

8 0

3 years ago