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3 years ago

Which inequality best represents that ice cream at −2°C is cooler than ice cream at 4°C?

Mathematics

2 answers:

Viefleur [7K]3 years ago

7 0

Answer:

your answer is 3°C > −2°C!

Step-by-step explanation:

I just took this test.

FLVS right, segment exam practice test?

Hop this helped fellow FLVS student!!!

IrinaK [193]3 years ago

4 0

Answer:

-2 C is cooler cause - in the C makes it cooler (sorry if im bad this is my 1st time)

Step-by-step explanation:

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Juan and his friends played a game in which sticky darts were thrown onto a playing board. To start the game, Juan threw a dart

evablogger [386]

Answer:

8 points: unlikely

16 points: impossible

more than 0 points: likely

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the surface area and volume of each figure. Round each answer to the nearest hundredth.

Llana [10]

Answer: Surface area is equal to 200 $cm^{2}$

Volume is equal to 333.33 $cm^{3}$

Step-by-step explanation:

First, let's do surface area.

The surface area of a pyramid is equal to 1/2(perimeter of base)(lateral height) + area of the base

The perimeter of the base is 10(4) = 40; as the base is a square with a side length of 10.

The lateral height is given as 5 cm.

The area of the base is 10(10) = 100.

We can plug those numbers into the equation to get 1/2(40)(5) + 100, which comes out to be 200 $cm^{2}$ .

Now for volume.

The volume of a pyramid is equal to 1/3(area of the base)(height).

We already have the area of the base, which is 100.

The height is given as 10 cm.

Plugging those numbers into the equation, we get 1/3(100)(10), which is 1000/3 or about 333.33 $cm^{3}$ .

Hope this helps!

6 0

3 years ago

The given triangles are similar. Solve for x.

Maru [420]

The first one is 21.93 rounded up to 22.

8 0

3 years ago

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p

Alekssandra [29.7K]

Using the z-distribution and the formula for the margin of error, it is found that:

a) A sample size of 54 is needed.

b) A sample size of 752 is needed.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence level, hence $\alpha = 0.9$ , z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$ , so $z = 1.645$ .

Item a:

The estimate is $\pi = 0.213 - 0.195 = 0.018$ .

The sample size is <u>n for which M = 0.03</u>, hence:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.018(0.982)}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.018(0.982)}$

$\sqrt{n} = \frac{1.645\sqrt{0.018(0.982)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.018(0.982)}}{0.03}\right)^2$

$n = 53.1$

Rounding up, a sample size of 54 is needed.

Item b:

No prior estimate, hence $\pi = 0.05$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.5(0.5)}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.645\sqrt{0.5(0.5)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.5(0.5)}}{0.03}\right)^2$

$n = 751.7$

Rounding up, a sample of 752 should be taken.

A similar problem is given at brainly.com/question/25694087

5 0

2 years ago

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

$A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]$

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$

$5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}$

6 0

3 years ago