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Sidana [21]
3 years ago
15

Solve each equation by taking square roots. Show your work. Simplify your answers. Do not write your answers in decimal form.

Mathematics
1 answer:
natulia [17]3 years ago
6 0

Step-by-step explanation:

\sf \hookrightarrow \:  {n}^{2}  = 20 \\ \sf \hookrightarrow \:  \sqrt{ {n}^{2} }  =  \sqrt{20}  \\ \sf \hookrightarrow \: n =  \sqrt{20}  \\  \star \bf \: n = 2 \sqrt{5}  \\  \\ \sf \hookrightarrow \:  {a}^{2}  = 90 \\ \sf \hookrightarrow \:  \sqrt{ {a}^{2} }  =  \sqrt{90}  \\ \sf \hookrightarrow \: a =  \sqrt{90}  \\  \star \bf \: a = 3 \sqrt{10}

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The fair spinner shown in the diagram above is spun work out the probability of getting a factor of 8​
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Answer:1/8

Step-by-step explanation:cuz

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4. A lampshade is in the shape of a cone. The radius iš 7 inches and the height 9
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Answer:

461.8

Step-by-step explanation:

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2 years ago
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Let x be a Poisson random variable with μ = 9.5. Find the probabilities for x using the Poisson formula. (Round your answers to
miv72 [106K]

Answer: 1

Step-by-step explanation: the probability mass function that defines a possion probability distribution is given below as

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P(x=0) = e^-0 × 9.5^0 / 0!

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3 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

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3 years ago
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AleksAgata [21]
It's 190 it's easy bro
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