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3 years ago

If f(x) = 7x2 - 4x, what is the value of f(2)? F -8 J 17 G 2 K 20 H 6

Mathematics

1 answer:

kow [346]3 years ago

4 0

Answer:

K 20

Step-by-step explanation:

f(x) = 7x2 - 4x

f (2) = 7 (2)^2 - 4(2)

= 28 - 8

= 20

Since they replaced the x inside f(x) with 2, just do that the same to the right side, change all x into 2

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Step-by-step explanation:

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The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are

almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = the number that are not correctly calibrated.

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_3$

So, the required probability = $\frac{^{6}C_3}{^{8}C_3}$

= $\frac{20}{56}$ = 0.35

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_2$

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_1$

So, the required probability = $\frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}$

= $\frac{30}{56}$ = 0.54

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_1$

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_2$

So, the required probability = $\frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}$

= $\frac{6}{56}$ = 0.11

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0

3 years ago

f is a differentiable function on the interval [0, 1] and g(x) = f(4x). The table below gives values of f '(x). What is the valu

polet [3.4K]

The table is attached in the figure.

g(x) = f(4x) ⇒⇒⇒ differentiating both sides with respect to x

∴ g'(x) = $\frac{d}{dx} [f(x)] * \frac{d}{dx} [4x]=4*f'(x)$ ⇒⇒⇒⇒⇒⇒ chain role

To find g '(0.1)

Substitute with x = 0.1

from table:

f'(0.1) = 1 ⇒ from the table

∴ g'(0.1) = 4 * [ f'(0.1) ] = 4 * 1 = 4

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3 years ago