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3 years ago

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What is the average rate of change of f(x), represented by the table of values, over the interval [-3, 4]? x f(x) -6 27 -3 6 -1

2 0 3 1 6 4 27 A. -6 B. -3 C. 3 D. 6 E. 21

1 answer:

USPshnik [31]3 years ago

3 0

Answer:

<em>The correct option will be: C. 3</em>

Step-by-step explanation:

The given table is..........

$x :$ -6 -3 -1 0 1 4

$f(x) :$ 27 6 2 3 6 27

<u>The formula for average rate of change</u> is: $\frac{f(b)-f(a)}{b-a}$ , where $[a,b]$ is the given interval.

Here the interval is [-3, 4]. So, $a=-3$ and $b=4$

Now, plugging the values of $a$ and $b$ into the above formula, we will get........

$\frac{f(4)-f(-3)}{4-(-3)}$

<u>From the given table,</u> we will get $f(4)=27$ and $f(-3)=6$

So, the average rate of change will be: $\frac{27-6}{4-(-3)}= \frac{21}{7}=3$

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WILL GIVE BRAINLIEST

galina1969 [7]

Answer:

x= 3 inches

Step-by-step explanation:

-The volume of a box is given by the formula:

$V=lwh\\\\l-length\\w-width\\h-height$

-We are given the dimensions h=x+2, l=2x+5 and w=4x-1.

We substitute this values in the formula and equate to the volume value.

$V=lwh\\\\605=(x+2)(2x+5)(4x-1)\\\\605=8x^3+34x^2+31x-10$

#We take the values to the same side and equate to zero;

$8x^3+34x^2+31x-615=0\\\\\\\#Factor\\\\\\(x-3)(8x^2+58x+205)=0$

#Applying the zero factor principal to obtain the different values of x:

$(x-3)=0\\\\\therefore x=3\ \ \ \ \ \ ...i\\\\(8x^2+58x+205)=0$

#We use the quadratic formula to solve the two other values of x:

$(8x^2+58x+205)=0\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\=\frac{-58\pm\sqrt{58^2-4\times 8\times 205}}{2\times 8}\\\\x=-3.625+3.533i \ and \ x=-3.625-3.533i$

The two other values are negatives. Ignore them since length cannot be a negative.

The only reasonable value of x is x=3

#We substitute in the formula to validate:

$V=lwh\\\\=(x+2)(2x+5)(4x-1)\\\\=(3+2)(2*3+5)(4*3-1)\\\\=605$

Hence, the value of x is 3 inches

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3 years ago

2y-10(y-3)=70<br><br> Help!! ASAP

kirill115 [55]

Answer:

y = -5

Step-by-step explanation:

2y - 10y + 30 = 70

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4 years ago

Read 2 more answers

The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) F

cricket20 [7]

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$n = 300, p = 0.02$

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

$E(X) = np = 300*0.02 = 6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42$

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040$

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023$

$P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143$

$P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436$

$P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485$

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

From c), we have that $P(X = 0) = 0.0023$ . So

$0.0023 + P(X \geq 1) = 1$

$P(X \geq 1) = 0.9977$

99.77% probability that we find at least 1 of the seniors are married

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