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4 years ago

Multiply. 9x/4b•2bx/3x^5. Simplify your answer as much as possible

Mathematics

1 answer:

stich3 [128]4 years ago

6 0

Answer:

3/2x^3

Step-by-step explanation:

multiply the numerators and the denominators

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Please help with these three questions and thank you

Advocard [28]

1.A

2.D

3.B

Step-by-step explanation:

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4 years ago

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Help me pleaseeeeeeee

LiRa [457]

Answer:

x = -4/3 or x = 3/2

Step-by-step explanation:

Let's solve your equation step-by-step.

6x^2−x−12=0

Step 1: Factor left side of equation.

(3x+4)(2x−3)=0

Step 2: Set factors equal to 0.

3x+4=0 or 2x−3=0

x = -4/3 or x = 3/2

Answer:

x = -4/3 or x = 3/2

5 0

2 years ago

How many terms are in the polynomial abcd + e - h 2?

Free_Kalibri [48]

There are three terms in this polynomial, making it a trinomial. The terms are "abcd", "e", and "h^2".

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4 years ago

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How do you write add 4 to n then multiply by 5 in algebraic expression

irina [24]

Here is the anwer to your question ,
n+4•5

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4 years ago

The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However,

horsena [70]

Answer:

The mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

Step-by-step explanation:

We are given that records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.

On a typical Saturday, the oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Assuming that data follows normal distribution.

Let $\bar X$ = sample mean time

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 time = 16.2 minutes

$\sigma$ = standard deviation = 3.4 minutes

n = sample size = 35 oil changes

Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;

P(X $\leq$ x) = 0.10 {where x is required mean oil-change time}

P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }$ ) = 0.10

P(Z $\leq$ $\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }$ ) = 0.10

Now, in the z table the critical value of X which represents the below 10% of the probability area is given as -1.282, that means;

$\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }$ = -1.282

x - 16.2 = $-1.282 \times {\frac{3.4}{\sqrt{35} } }$

x = 16.2 - 0.74 = 15.46 minutes

Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

8 0

3 years ago