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3 years ago

Please help :{

Mathematics

2 answers:

faltersainse [42]3 years ago

3 0

Answer: She can buy 6 shirts

Step-by-step explanation: You subtract 95 from 240, which gives you 145 so you do 145/22.5 which will give you 6.4444 but you can't buy .44 of a shirt so you round down and you get 6.

andrezito [222]3 years ago

3 0

She can get 6 shirts

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10. Choose all fractions equivalent to 36/40 <br><br><br><br> A) 3/4 B) 9/10 C) 9/40 D) 18/20

Talja [164]

Answer:

B, D

Step-by-step explanation:

36/40 simplified is 9/10

9/10 x 2 = 18/20

I hope this helps!

5 0

3 years ago

Read 2 more answers

How do I solve this equation 9-x/5=3

tatiyna

Answer:

x= -10

Step-by-step explanation:.......

4 0

3 years ago

Directions: Calculate the area of a circle using 3.14x the radius

Leokris [45]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

$\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 8.41$

$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 1 \div 2$

$\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd$

Now, let's calculate area ~

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 0.25$

$\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 64$

$\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2}$

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8 0

3 years ago

How many quarts of pure antifreeze must be added to 8 quarts of a 40% antifreeze solution to obtain a 60% antifreeze solution

anyanavicka [17]

Let <em>q</em> be the number of quarts of pure antifreeze that needs to be added to get the desired solution.

8 quarts of 40% solution contains 0.40 × 8 = 3.2 quarts of antifreeze.

The new solution would have a total volume of 8 + <em>q</em> quarts, and it would contain a total amount of 3.2 + <em>q</em> quarts of antifreeze. You want to end up with a concentration of 60% antifreeze, which means

(3.2 + <em>q</em>) / (8 + <em>q</em>) = 0.60

Solve for <em>q</em> :

3.2 + <em>q</em> = 0.60 (8 + <em>q</em>)

3.2 + <em>q</em> = 4.8 + 0.6<em>q</em>

0.4<em>q</em> = 1.6

<em>q</em> = 4

4 0

3 years ago

A psychologist obtains a random sample of 2020 mothers in the first trimester of their pregnancy. The mothers are asked to play

leva [86]

Answer:

First of all, we need to determine the system of hypothesis, which must have a null and alternative hypothesis.

In all hypothesis proof procedure, we must work with the null one, rejecting it or accepting it. If we reject the null hypothesis, it will mean that our principal hypothesis is true.

According to the problem, the null hypothesis must be: $H_{o} : u = 100$ ; because it says that 100 is the normal mean that is used.

The alternative hypothesis must be: $H_{a} : u > 100$ ; because the question is asking if there's enough evidence to say that they children have a higher IQ than 100.

Then, we must calculate the P-value, which is the probability value that we are gonna use to accept or reject the null hypothesis. This P-value we can calculate it with the formula: $Z=\frac{x-u}{\frac{o}{\sqrt{n} } }$ ; where x is the sample mean, u is the hypothetical mean parameter, o is the sample deviation and n is the sample.

Extracting all values from the problem: x = 104.3; u = 100; o = 14 and n = 20.

Applying all values to the formula: $Z=\frac{104.3 - 100}{\frac{14}{\sqrt{20} } } = \frac{4.3}{\frac{14}{ 4.5} }=\frac{4.3}{3.1}=1.4$

Then, this Z-value will give us the P-value by using the Z-table with the 0.05 level of significance. In table attached, we can see that the result is 0.9265. However, the P-value is 1 - 0.9265 = 0.735. We need to do this subtraction because, the result 0.9265 belong to the bigger area (as it's shown is the image in blue colour), and we need the smaller area, which belong to the 0.05 level of significance.

<em>In this problems, if the P-value is more than the level of significance, the null hypothesis is accepted. If the P-value is less than the level of significance the null hypothesis is rejected.</em>

Therefore, there is no enough evidence to say that the sample children have higher IQs, because the P-value is higher than the level of significance. (0.735>0.05).

5 0

3 years ago