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Vlad [161]
2 years ago
11

Which inequality written in vertex form represents the graphed region?

Mathematics
1 answer:
DerKrebs [107]2 years ago
7 0

Answer:

C

Step-by-step explanation:

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In a standard normal distribution, what z value corresponds to 17% of the data between the mean and the z value?
hjlf
You're looking for a value z such that

\mathbb P(0

Because the distribution is symmetric, the value of z in either case will be the same.

Now, because the distribution is continuous, you have that

0.17=\mathbb P(0

The mean for the standard normal distribution is 0, and because the distribution is symmetric about its mean, it follows that \mathbb P(Z.

0.17=\mathbb P(Z

You can consult a z score table to find the corresponding score for this probability. It turns out to be z\approx0.4399.
4 0
3 years ago
Cheryl can travel 21.6 miles on one gallon of gas. how far can cheryl travel on 6.2 gallons of gas?
boyakko [2]
21.6 miles per gallon

Multiplied by 6.2 gallons 

21.6 * 6.2 = 133.92

Cheryl can travel 133.92 (or about 134) miles on 6.2 gal. of gas.

Hope that helped!
6 0
3 years ago
The sum of the speed of two trains is 722.7 miles per hour. If the speed of the first train is 3.3 mph faster than the second tr
stellarik [79]
x-speed\ of\ first\ train\\x-3.3-speed\ of\ second\ train\\\\x+x-3.3=722.7\\x+x=722.7+3.3\\2x=726\\x=363\\\\x-3.3=363-3.3=359.7\\\\Speed\ of\ first\ train\ is\ 363 mph,\ speed\ of\ second\ train\ is\ 359.7.
7 0
3 years ago
Joni has a circular garden with a diameter of 14 1/2 feet if she uses 2 teaspoons of fertilizer for every 25 square feet of gard
alina1380 [7]

Answer:

13.20 teaspoon of fertiliser needed for entire garden.

Step-by-step explanation:

Diameter of the garden = 14.50 ft

Radius of the garden = 7.25 ft

Area

[tex]A=\pi r^2[/text]

[tex]A=\pi \times 7.25 \times 7.25[/text]

[tex]\pi=3.14[/text]

[tex]A=\pi \times 7.25 \times 7.25[/text]

[tex]A=3.14 \times 7.25 \times 7.25[/text]

Hence

[tex]A=165.046[/text]

Area is 165.046 Sq Ft

For 25 sq ft of fields the fertiliser needed = 2 tea spoon

For 1 sq ft of fields the fertiliser needed =

[tex]\frac{2}{25}[/text] tea spoon

For 165.046 sq ft of fields the fertiliser needed =

[tex]\frac{2 \times 165.046 }{25}[/text] tea spoon

For 165.046 sq ft of fields the fertiliser needed = 13.20 tea spoon

6 0
3 years ago
F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))
Marrrta [24]

Given:

f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6

Required:

We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.

Explanation:

Given equation is

F(x)=\sqrt[4]{f(x)g(x)}.F(x)=(f(x)g(x))^{\frac{1}{4}}F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}

Differentiate the given equation for x.

Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{  Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.

F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)

Replace x=3 in the equation.

F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}F^{\prime}(3)=\frac{-6-1}{4}F^{\prime}(3)=\frac{-7}{4}

Final answer:

F^{\prime}(3)=\frac{-7}{4}

8 0
1 year ago
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