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2 years ago

186% of what is 56.73

1 answer:

4 0

To do this problem, first you are going to want to turn 186% into a decimal which is 1.86. Now you put 56.73 over 1.86.................. 56.73/1.86.
Then you should divide 56.73 by 1.86 then you should get 30.5.

EXTRA.
if you want to find the percent, all you have to do is multiply 30.5 by 100.
you would get 3050%
Hope this helped and have an awesome day!

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Poiseuille's Law states that the volume, V, of blood flowing through an artery in a unit of time at a fixed pressure is directly

eimsori [14]

Answer:

11.58%

Step-by-step explanation:

The initial volume if blood flowing through the artery is given by

$V=kr^4$

To achieve a new volume of 155% (55% increase) of the initial volume, the new radius must be:

$V'= 1.55V\\1.55V=k(r')^4\\1.55kr^4 = k(r')^4\\(\sqrt[4]{1.55}*r)^4=(r')^4 \\(1.1158*r)^4=(r')^4 \\r'=1.1158*r$

Since the new radius is 1.1158 times larger than the initial radius, the percentage increased is:

$I=(1.1158 - 1)*100\%\\I= 11.58\%$

7 0

2 years ago

Ali, Ben and Clare each played a game. Clare's score was seven times Ali's score. Ben's score was half of Clare's score. Write d

Lerok [7]

Answer:

The answer is 2 : 7 : 14

8 0

2 years ago

A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well

kogti [31]

Answer:

Step-by-step explanation:

Hello!

To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:

<u>Group 1</u>: Ate healthy breakfast

X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.

n₁= 26

<u>Group 2: </u>Did not eat healthy breakfast

X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.

n₂= 20

After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:

p'₁= 0.50

p'₂= 0.40

The parameters of study are the population proportions, if the claim is true then p₁ > p₂

And you can determine the hypotheses as

H₀: p₁ ≤ p₂

H₁: p₁ > p₂

α: 0.05

$Z= \frac{(p'_1-p'_2)-(p_1-p_2)}{\sqrt{p'(1-p')[\frac{1}{n_1} +\frac{1}{n_2}] } } }$ ≈N(0;1)

pooled sample proportion: $p'= \frac{x_1+x_2}{n_1+n_2} =\frac{13+8}{46} = 0.46$

$Z_{H_0}= \frac{(0.5-0.4)-0}{\sqrt{0.46(1-0.46)[\frac{1}{26} +\frac{1}{20}] } } }= 0.67$

p-value: 0.2514

The decision rule is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.

At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.

So having a healthy breakfast doesn't seem to improve the grades of students.

I hope this helps!

5 0

2 years ago

A student walks 1/4 Mile from her home to the store on her way to a friend's house if the store is 1/3 of the way to a friend's

exis [7]

the store is 1/3 the way to her friend's house, not 1/3 mile.

this explains solution:

1/3x = 1/4

x = 1/4 * 1/3

x = 3/4

Notes:

X: is the total distance.

1/3 is the one portion of total distance.

1/4 mile is the distance store.

3 0

2 years ago

A basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modele

tresset_1 [31]

Answer:

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

Step-by-step explanation:

$\text{The equation for ball height} : 6+30\cdot t-16\cdot t^{2}\\\text{The equation for shot blocker's height} : 9+25\cdot t-16\cdot t^{2}$

But, the shot is made before two tenths of a second or 0.2 seconds therefore modified equation for ball height is :

$6+30\cdot (t-0.2)-16\cdot (t-0.2)^{2}$

Now for the shot to be blocked,the height of shot blocker must be greater than the height of the ball which is shot before 0.2 seconds :

$\implies 9+25\cdot t-16\cdot t^{2}\geq 6+30\cdot (t-0.2)-16\cdot (t-0.2)^{2}\\\implies 9+25\cdot t-16\cdot t^{2}\geq 6+30\cdot t-6-16\cdot (t^{2}-0.4\cdot t+0.04)\\\implies9+25\cdot t-16\cdot t^{2}\geq 30\cdot t-16\cdot t^{2}+6.4\cdot t-0.64\\\implies 9+25\cdot t\geq 36.4\cdot t-0.64\\\implies 9.64\geq 11.4\cdot t\\\\\implies t\leq \frac{9.64}{11.4}\approx 0.846$

Hence, the shot was blocked between 0.84 and 0.85 seconds after the shot is launched.

7 0

2 years ago

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