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shtirl [24]
3 years ago
8

What is the surface area of this

Mathematics
2 answers:
dem82 [27]3 years ago
7 0
The answer is 95 
Multiply l*w*h 
5*3*4
95
kiruha [24]3 years ago
3 0
The area would be 94m
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Solve the formula V=pir^2h for r <br><br> PLEAASSSEEE HELP
otez555 [7]

Answer:

B

Step-by-step explanation:

So we have the formula:

V=\pi r^2h

And we want to solve it for r.

So, let's first divide both sides by π and h. This will cancel out the right side:

r^2=\frac{V}{\pi h}

Now, take the square root of both sides:

r=\sqrt{\frac{V}{\pi h}}

And we're done!

Our answer is B.

I hope this helps!

7 0
3 years ago
Read 2 more answers
Need help can someone help me pleaseeee
Hunter-Best [27]

at first we write each one as a sequence

and find its equation

the first table

-19, -11, -3, 5

8x-19

the third table

15,12,9,6

-3x+15

the second table

-1. 5,1.5,3,4.5

+3 +1.5 +1.5

so the third table is the nonlinear function

8 0
3 years ago
Show that there is no solution for the radical equation 4w+4=-4.
Ad libitum [116K]

Answer:

there is a solution

w= -2

Step-by-step explanation:

4w+4=-4

    -4    -4

4w=-8

/4    /4

w=-2

3 0
3 years ago
One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo
almond37 [142]

Answer:

(1) P(\bar X < 26 inches) = 0.0436

(2) P(27.5 inches < \bar X < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let \bar X = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

            Z = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean vertical  leap = 28 inches

            \sigma = standard deviation = 7 inches

            n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P(\bar X < 26 inches)

   P(\bar X < 26) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{26-28}{\frac{7}{\sqrt{36} } } ) = P(Z < -1.71) = 1 - P(Z \leq 1.71)

                                                 = 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < \bar X < 28.5 inches) = P(\bar X < 28.5 inches) - P(\bar X \leq 27.5 inches)

    P(\bar X < 28.5) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{28.5-28}{\frac{7}{\sqrt{26} } } ) = P(Z < 0.36) = 0.64058 {using z table}                      

    P(\bar X \leq 27.5) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{27.5-28}{\frac{7}{\sqrt{26} } } ) = P(Z \leq -0.36) = 1 - P(Z < 0.36)

                                                        = 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < \bar X < 28.5 inches) = 0.64058 - 0.35942 = 0.2812

6 0
3 years ago
Someone answer this quick
Gemiola [76]

Answer:

Step-by-step explanation:

if im not wrong it should be the first one

5 0
1 year ago
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