Ask question

Ask question

All categories

3 years ago

12

In Tim's math class, there are 12 pencils and 15 markers. Which of the following is the ratio of markers to pencils in Tim's mat

h class?

1 answer:

Hunter-Best [27]3 years ago

6 0

Answer:

3

Step-by-step explanation:

there are 3 more markers than pencils

You might be interested in

Will mark the brainliest Given the following triangle with the indicated height and base.

Serjik [45]

POST A PHOTO ZOOMED OUT SO I CAN HELP THANKS

5 0

3 years ago

Untitled Question

Lunna [17]

I think the answer is D if not sorry

4 0

3 years ago

A recent national survey found that high school students watched an average (mean) of 7.1 movies per month with a population sta

muminat

Answer:

H0: μc ≤ μs Ha :μc > μs

Step-by-step explanation:

The null and alternate hypotheses can be stated as

H0: μc ≤ μs Ha :μc > μs one tailed test

Where

μc = Mean of college students watching movies in a month

μs = Mean of school students watching movies in a month

For one tailed test of α =0.05 the value of Z= ± 1.645

The critical region will be Z > ± 1.645

It is of importance to note that by rejecting the null hypothesis and accepting the alternate hypothesis we are automatically rejecting all values of mean that are greater than 7.1

3 0

3 years ago

What is the purpose of sampling?

Kryger [21]

Answer:

A. to make inferences about a sampling technique

Step-by-step explanation:

6 0

3 years ago

Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.

rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed as:

$y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}$

Thus, the area of the surface is as follows:

$\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA$

$= \iint_D \sqrt{x^2 +y^2 +1 } \ dA$

$= \int^{2 \pi}_{0} \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr \ d \theta$

$=2 \pi \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr$

$= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0$

$= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg]$

$= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]$

3 0

3 years ago