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mojhsa [17]
3 years ago
7

Look at each expression below. Is it equivalent to 42x - 56y? Select Yes or No for a. 40(2x-16y)

Mathematics
1 answer:
MrRa [10]3 years ago
3 0

Answer:

A. no

Step-by-step explanation:

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Assume that human body temperatures are normally distributed with a mean of 98.22 degrees Upper F 98.22°F and a standard deviati
Anna35 [415]

Answer: 0.00102%

Step-by-step explanation:

Given : Human body temperatures are normally distributed with a mean of \mu=98^{\circ}Fand a standard deviation of s=0.61^{\circ}F

A hospital uses 100.6^{\circ}F as the lowest temperature considered to be a fever.

Let x be the random variable that represents the human body temperatures.

z=\dfrac{x-\mu}{s}

For x= 100.6, z=\dfrac{100.6-98}{0.61}=4.26229508197\approx4.26

Using normal distribution table for z-values for right-tailed area ,

P(x>100.6)=P(Z>4.26)=0.0000102=0.00102\%

Hence, the required probability = 0.00102%

8 0
3 years ago
Jonah earns $12.50 per hour as a marketing intern. If he works for 4.5 hours,how many dollars does he earn?
Margarita [4]
$56.25. Just take a calculator and write $12.50 times * 4.5 equals=.
5 0
3 years ago
What percent of 200 is 0.5?
OleMash [197]
<span>Question: What percent of 200 is 0.5?
Solutions and discussions:
=> 200 is the 100% value of the numbers
=> 0.5 is the number in which we need to find the percentage value of it in 200.
How to solve. Simply follow the formula
=> 0.5 / 200 = 0.0025 – this is the to get the decimal value of the number that we will be converting to percentage.
=> 0.0025 * 100% = 0.25%
Therefore 0.5 is 0.25% of 200.

</span>



3 0
4 years ago
Read 2 more answers
which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater
Nutka1998 [239]

Answer:

\frac{\sqrt[3]{16y^4}}{x^2}

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices \sqrt[n]{a}  = a^{\frac{1}{n}}

So,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } gives

\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}

Also from laws of indices

(ab)^n = a^nb^n

So, the above expression can be further simplified to

\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}

Multiply the exponents gives

\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

Substitute 2^5 for 32

\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

From laws of indices

\frac{a^m}{a^n} = a^{m-n}

This law can be applied to the expression above;

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})} becomes

2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}

Solve exponents

2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}

From laws of indices,

a^{-n} = \frac{1}{a^n}; So,

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}} gives

\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}

The expression at the numerator can be combined to give

\frac{(2y)^{\frac{4}{3}}}{x^2}

Lastly, From laws of indices,

a^{\frac{m}{n} = \sqrt[n]{a^m}; So,

\frac{(2y)^{\frac{4}{3}}}{x^2} becomes

\frac{\sqrt[3]{(2y)}^{4}}{x^2}

\frac{\sqrt[3]{16y^4}}{x^2}

Hence,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } is equivalent to \frac{\sqrt[3]{16y^4}}{x^2}

8 0
3 years ago
On Wednesday,Felipe drove 150 miles. This was 3 times the he distance he drove on Monday and 40 miles more than they drove on Tu
insens350 [35]
36 take 150-40 then you would want to decide 3 from 110
8 0
3 years ago
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