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4 years ago

1) If f(x) = 3x2 – Bx + 4 and f(-1) = 12, what is the value of B?

Mathematics

1 answer:

iren2701 [21]4 years ago

5 0

Answer:

$B=-5$

Step-by-step explanation:

we have

$f(x)=3x^{2} -Bx+4$

we know that

f(-1) is the value of the function for x=-1

substitute the value of x=-1 in the function and solve for B

$f(-1)=3(-1)^{2} -B(-1)+4$

we have

$f(-1)=12$

substitute

$12=3(-1)^{2} -B(-1)+4$

$12=3-B+4$

$12=7-B$

$B=7-12$

$B=-5$

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The mean incubation time of fertilized chicken eggs kept at 100.5 degree F in a still-air incubator is 21 days. Suppose that the

Nitella [24]

Given Information:

Mean incubation time = 21 days

Standard deviation of incubation time = 1 day

Required Information:

a) P(X < 20) = ?

b) P(X > 22) = ?

c) P(19 < X < 21) = ?

Answer:

a) P(X < 20) = 15.87

b) P(X > 22) = 15.87

c) P(19 < X < 21) = 47.23%

Explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

a) We want to find out the probability that a randomly selected fertilized chicken egg hatches in less than 20 days.

$P(X < 20) = P(Z < \frac{x - \mu}{\sigma} )\\\\P(X < 20) = P(Z < \frac{20 - 21}{1} )\\\\P(X < 20) = P(Z < \frac{-1}{1} )\\\\P(X < 20) = P(Z < -1)\\$

The z-score corresponding to -1 is 0.1587

$P(X < 20) = 0.1587\\\\P(X < 20) = 15.87 \%$

Therefore, the probability that a randomly selected fertilized chicken egg hatches in less than 20 days is 15.87%

b) We want to find out the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch?

$P(X > 22) = 1 - P(X < 22)\\\\P(X > 22) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\P(X > 22) = 1 - P(Z < \frac{22 - 21}{1} )\\\\P(X > 22) = 1 - P(Z < \frac{1}{1} )\\\\P(X > 22) = 1 - P(Z < 1)\\$

The z-score corresponding to 1 is 0.8413

$P(X > 22) = 1 - 0.8413\\\\P(X > 22) = 0.1587\\\\P(X > 22) = 15.87 \%$

Therefore, the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch is 15.87%

c) We want to find out the probability that a randomly selected fertilized chicken egg hatches between 19 and 21 days?

$P(19 < X < 21) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(19 < X < 21) = P( \frac{19-21}{1} < Z < \frac{21 - 21}{1} )\\\\P(19 < X < 21) = P( \frac{-2}{1} < Z < \frac{0}{1} )\\\\P(19 < X < 21) = P( -2 < Z < 0 )\\$

The z-score corresponding to -2 is 0.0227 and 0 is 0.50

$P(19 < X < 21) = P( Z < 0 ) - P( Z < -2 ) \\\\P(19 < X < 21) = 0.50 - 0.0227 \\\\P(19 < X < 21) = 0.4723\\\\P(19 < X < 21) = 47.23 \%$

Therefore, the probability that a child spends more than 4 hours and less than 8 hours per day unsupervised is 47.23%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.0, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.00 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

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4 years ago

Translate each phrase into a variable expression. use n for the variable

gayaneshka [121]

N + 2 that is correct

8 0

4 years ago

Read 2 more answers

One pipe can empty a tank 5 times faster than another pipe can fill the tank. Starting with a full tank, if both pipes are turne

xxTIMURxx [149]

Answer:

40 hours

Step-by-step explanation:

Let's say that the slower pipe can fill x amount of the tank in one hour. It adds x amount to the tank every hour. Therefore, we can say, for the slower pipe,

1 hour = x amount

Then, the faster pipe empties the tank 5 times faster than the slower pipe fills it, so it removes 5 times the amount that the smaller pipe puts in, so for the faster pipe,

1 hour = -5x amount (negative to symbolize removing).

For the problem at hand, the tank starts at full, or 100%=1. It ends empty, or at 0% = 0. After 10 hours, if we only account for the slower pipe, we add x amount to the tank every hour, so we add 10 times that total, resulting in

1 + 10 * x as the ending result if we don't include the faster pipe. Then, the faster pipe removes 5x every hour, so in 10 hours, it removes 50x, so we have

1+10 * x - 50 * x as the final amount of stuff in the tank, which is equal to 0. Therefore, we have

1 + 10 * x - 50 * x = 0

1 - 40 * x = 0

add 40*x to both sides to isolate the x and its coefficient

1 = 40 * x

divide both sides by 40 to isolate x

x= 0.025

Therefore, the slower pipe adds 0.025, or 1/40 = 2.5% to the tank every hour. We want to figure out how long it would take for the slower pipe to fill up an empty tank, or turn it from 0% to 100% full.

Because the slower pipe adds 2.5% of the tank every hour, we can say that over y hours, it fills up

2.5% * y amount of the tank. We want to figure out how many hours it would take to make it 100% (we need to add 100% of the tank in the problem), so we can say

2.5% * y = 100%

divide both sides by 2.5% to isolate y

100%/2.5% = y = 40

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