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Leya [2.2K]
3 years ago
5

What are all the values of x for which the function f defined by f(x) = (x^2 - 3)e^-xis increasing?

Mathematics
1 answer:
Goryan [66]3 years ago
8 0
I’m pretty sure it’s D
You might be interested in
Can anyone help me please
Ksju [112]
Pemdas

in the parenthasees, we have 2*3 which is 6
then 6+21=27
now we have
(1/3)(27) times 3^2-5.6
exponents,, 3^2=9

(1/3)(27)(9)-5.6
multiply
81-5.6
minus
75.4
5 0
4 years ago
Please help and thank you
Black_prince [1.1K]

I'm honestly not really sure, but it has to be either (2,10) or (10,2). I've never seen a question like this...

Answer:

Step-by-step explanation:

8 0
4 years ago
What is 1) 0.036 2) 0.099 3) 0.865 as a fraction with a denominator of 20?
Andrews [41]

Answer:

(1.) 0.72/20  (2.) 1.98/20 (3.) 17.3/20

Step-by-step explanation:

Multiply .036, .099, and .865 each indivigually by 20 to get the numerators for each fraction.

4 0
3 years ago
Pls help me...........
schepotkina [342]

Answer:

\frac{4}{5}

Step-by-step explanation:

Using the Pythagorean identity

sin²x + cos²x = 1, then

sin²x = 1 - cos²x

sinx = \sqrt{1-cos^2x}

note that cosΘ = \frac{6}{10} = \frac{3}{5}

sinΘ = \sqrt{1-(3/5)^2}

        = \sqrt{1-\frac{9}{25} }

        = \sqrt{\frac{16}{25} } = \frac{4}{5}

6 0
3 years ago
In a batch of 100 cell phones, there are, on average, 5 defective ones. if a random sample of 30 is selected, find the probabili
Dafna11 [192]

Let x be the number of defective cell phones. It is given that in batch of 100, on average 5 are defective.

let p be the probability of defective cell phone.

p = 5/100 = 0.05

Let n be size of random sample, n=30

Here out of 30 we want to find probability that 2 will be defective. It means 30-2 =28 cell phones will be non defective.

The probability of getting non defective cell phone is 1- p=1-0.05 =0.95

The probabability of getting 2 defective is

P(X=2) = number of ways selecting 2 from 30 * probability 2 defective * probability of 28 non defective

Now number of ways of selecting 2 cell phone from 30 is

30C2 = \frac{30!}{(30-2)! 2!}

= \frac{30!}{28! 2!}

= \frac{30 *29* 28!}{28! 2!}

= (30*29) /2

30C2 = 435

P(X=2) = 30C2 * (0.05)^2 * (0.95)^28

= 435 * 0.0025 * 0.2378

P(X=2) = 0.2586

Probability of getting 2 defective out of 30 is 0.2586

5 0
4 years ago
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