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2 years ago

What is the product of (2p + 7)(3p2 + 4p – 3)?

Mathematics

1 answer:

Nady [450]2 years ago

3 0

Answer:

(2p+7)(3p²+4p-3)

2p(3p²+4p-3)+7(3p²+4p+3)

(6p³+8p²-6p)+ (21p²+28p+21)

6p³+29p²-6p+21

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I just need help with 14,15,and the bonus

ira [324]

Answer:

9: a^8/c^2

Step-by-step explanation:

(a^4/c)^2

(a^8/c^2)

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2 years ago

What are the solutions to the system of equations graphed below? Select all

saul85 [17]

Answer:

c and d

Step-by-step explanation:

the x intercepts are the solutions

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3 years ago

What is the midline equation for the function h(x)=−3cos(πx+2)−6?

Rus_ich [418]

Answer:

y = -6

Step-by-step explanation:

y=acos(bx+c)+d

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3 years ago

Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.

lesya [120]

Answer:

0.1994 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851$

P(sample of 20 boxers is more than 167 pounds)

$P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)$

$= 1 - P(z \leq 0.8438)$

Calculation the value from standard normal z table, we have,

$P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%$

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0

3 years ago

Please help me out! It’s urgent!

LuckyWell [14K]

Answer:

8 square units

Step-by-step explanation:

Area of the shaded shape = sum of shaded squares × 1 square unit

= 8 × 1 = 8 square units

6 0

2 years ago