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3 years ago

What is the point of maximum growth:

Mathematics

1 answer:

salantis [7]3 years ago

7 0

f(h(x))= 2x -21

Step-by-step explanation:

f(x)= x^3 - 6

h(x)=\sqrt[3]{2x-15}

WE need to find f(h(x)), use composition of functions

Plug in h(x)

f(h(x))=f(\sqrt[3]{2x-15})

Now we plug in f(x) in f(x)

f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6

cube and cube root will get cancelled

f(h(x))= 2x-15 -6= 2 x-21

wait i think i did the wrong one brb

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Simplify the expression 4(x − 5)(x2 + x + 2).

N76 [4]

Answer:

4x^3 − 16x^2 − 12x − 40

Step-by-step explanation:

= 4(x)(x^2 +x +2) +4(-5)(x^2 +x +2) . . . use the distributive property

= 4x^3 +4x^2 +8x -20x^2 -20x -40 . . . . and again

= 4x^3 -16x^2 -12x -40

3 0

3 years ago

Determine if the statement is true or false. If a linear system has the same number of equations and variables, then it must hav

lutik1710 [3]

Answer:

False

Step-by-step explanation:

Consider the equations with the same number of equations and variables as shown below,

$x_{1} + x_{2} = 0\\x_{1} + x_{2} = 1$

This equation has no solution because it is not possible to have two numbers that give a sum of 0 and 1 simultaneously.

$x_{1} + x_{2} = 1\\2x_{1} + 2x_{2} = 2$

This equation has infinitely many possible solutions.

Therefore it is FALSE to say a linear system with the same number of equations and variables, must have a unique solution.

5 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago

On a particular production line,the likelihood that a light bulb is defective is 5%. Ten light bulbs are randomly selected. What

KengaRu [80]

Answer:

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

Step-by-step explanation:

Consider the provided information,

Let X is the number of defective bulbs.

Ten light bulbs are randomly selected.

The likelihood that a light bulb is defective is 5%.

Therefore sample size is = n = 10

Probability of a defective bulb = p = 0.05.

Therefore, q = 1 - p = 1 - 0.05 = 0.95

Mean of binomial random variable: $\mu=np$

Therefore, $\mu=10(0.05)=0.5$

Variance of binomial random variable: $\sigma^2=npq$

Therefore, $\sigma^2=10(0.05)(0.95)=0.475$

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

5 0

3 years ago

Assume all the variables are nonnegative.

inn [45]

Use photo math it’s a great tool

6 0

2 years ago