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topjm [15]
4 years ago
12

FUNCTIONS: In the space provided, type the answer in descending order as it applies without any spaces between the letters, numb

ers, or symbols.
Type the composition (fog)(x) of the given functions:
f(x) = x^2 + 2x − 6 and g(x) = x + 5.
Mathematics
1 answer:
Aleonysh [2.5K]4 years ago
3 0

Answer:

Hence The composition (fog)(x) of the given function is x^2+12x+29.

Step-by-step explanation:

Given:

f(x) = x^2+2x-6

g(x)=x+5

We need to find (f o g)(x).

Solution:

Now we can say that;

(f o g)(x) = f(g(x))

(fog)(x) = (x+5)^2+2(x+5)-6

Now Applying distributive property we get;

(fog)(x) = (x+5)^2+2\times x+2\times5-6\\\\(fog)(x) = (x+5)^2+2x+10-6\\\\(fog)(x) = (x+5)^2+2x+4

Now Solving the exponent function we get;

(fog)(x) = x^2+2\times x\times 5+5^2+2x+4\\\\(fog)(x) = x^2+10x+25+2x+4\\\\(fog)(x) = x^2+12x+29

Hence The composition (fog)(x) of the given function is x^2+12x+29.

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Step-by-step explanation:

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4 years ago
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A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a samp
fenix001 [56]

Answer:

The lower bound of a 99% C.I for the proportion of defectives = 0.422

Step-by-step explanation:

From the given information:

The point estimate = sample proportion \hat p

\hat p = \dfrac{x}{n}

\hat p = \dfrac{55}{100}

\hat p = 0.55

At Confidence interval of 99%, the level of significance = 1 - 0.99

= 0.01

Z_{\alpha/2} =Z_{0.01/2} \\ \\ = Z_{0.005} = 2.576

Then the margin of error E = Z_{\alpha/2} \times \sqrt{\dfrac{\hat p(1-\hat p)}{n}}

E = 2.576 \times \sqrt{\dfrac{0.55(1-0.55)}{100}}

E = 2.576 \times \sqrt{\dfrac{0.2475}{100}}

E = 2.576 \times0.04975

E = 0.128156

E ≅ 0.128

At 99% C.I for the population proportion p is: \hat p - E

= 0.55 - 0.128

= 0.422

Thus, the lower bound of a 99% C.I for the proportion of defectives = 0.422

6 0
3 years ago
1. Suppose we have a six-sided die that we roll once. Let ai represent the event that the result is i. Let Bj represent the even
Alex787 [66]

Using the probability concept, it is found that there is a

a) 0.2 = 20% probability that 3 is obtained.

b) 0.3333 = 33.33% probability that 6 is obtained.

c) 0.6667 = 66.67% probability of a number greater than 3.

d) 0.6667 = 66.67% probability of an even number.

A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.

Item a:

There are 5 numbers greater than 1, one of which is 3, hence:

p = \frac{1}{5} = 0.2

0.2 = 20% probability that 3 is obtained.

Item b:

There are 3 numbers greater than 3, one of which is 6, hence:

p = \frac{1}{3} = 0.3333

0.3333 = 33.33% probability that 6 is obtained.

Item c:

There are 3 even numbers, two of which are greater than 3, hence:

p = \frac{2}{3} = 0.6667

0.6667 = 66.67% probability of a number greater than 3.

Item d:

There are 3 numbers greater than 3, two of which are even, hence:

p = \frac{2}{3} = 0.6667

0.6667 = 66.67% probability of an even number.

A similar problem is given at brainly.com/question/25667645

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