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3 years ago

10

Divide. (

iddle" class="latex-formula"> + 12x + 36) ÷ ( $x^{2} x^{2}$ + 11x + 30)

1 answer:

Maksim231197 [3]3 years ago

6 0

(x+6)^2
----------------
x^4+11x+30

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IceJOKER [234]

Answer:

21%

Step-by-step explanation:

3*7/42*3=21/100=21%

6 0

2 years ago

Max <br> p=3x+2y <br> subject to <br> 5x+y<16 <br> 2x+3y<22 <br> x>0 <br> y>0

garri49 [273]

Answer:For all the corner points, the maximum is at point (1, 3)

From the graph of the constraints, the corner points of the feasibility region are (0, 0), (0, 10/3), (1, 3), (2, 0)

For (0, 0): p = 0 + 2(0) = 0

For (0, 10/3): p = 0 + 2(10/3) = 20/3 = 6.67

For (1, 3): p = 1 + 2(3) = 1 + 6 = 7

For (2, 0): p = 2 + 2(0) = 2

Therefore, solution = (1, 3)

3 0

3 years ago

If f(x) = 56-2x, find f(7)

Lorico [155]

Answer: 42

Step-by-step explanation:

Simply substitute 7 for x.

56 - 2(7)

56 - 14

42

Hope it helps <3

3 0

3 years ago

Read 2 more answers

Frank has a 2-digit number on his baseball uniform. The number is a multiple of 10 and has 3 for one of its factors . What three

White raven [17]

Frank could have 30, 60 , 90 on his uniform

<em><u>Solution:</u></em>

Given that, Frank has a 2-digit number on his baseball uniform

The number is a multiple of 10 and has 3 for one of its factors

To find: Three numbers that Frank have on his uniform

Given that,

It's a multiple 10 and has 3 for one of its factors and its a 2-digit number

Let us first find the multiples of 10

<em><u>The two digit multiples of 10 are:</u></em>

{10, 20, 30, 40, 50, 60, 70, 80, 90}

From the above list, find the numbers which has 3 as one of its factors

Which means, find the numbers which are divisible by 3

The possible numbers = 30, 60, 90

Thus frank could have 30, 60 , 90 on his uniform

3 0

3 years ago

The average American man consumes 9.8 grams of sodium each day. Suppose that the sodium consumption of American men is normally

Alex Ar [27]

Answer:

(a) The distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).

(b) The probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c) The middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

Step-by-step explanation:

The random variable <em>X</em> is defined as the amount of sodium consumed.

The random variable <em>X</em> has an average value of, <em>μ</em> = 9.8 grams.

The standard deviation of <em>X</em> is, <em>σ</em> = 0.8 grams.

(a)

It is provided that the sodium consumption of American men is normally distributed.

The random variable <em>X</em> follows a normal distribution with parameters <em>μ</em> = 9.8 grams and <em>σ</em> = 0.8 grams.

Thus, the distribution of <em>X</em> is <em>N</em> (9.8, 0.8²).

(b)

If X ~ N (µ, σ²), then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).

To compute the probability of Normal distribution it is better to first convert the raw score (<em>X</em>) to <em>z</em>-scores.

Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:

$P(8.8$

$=P(-1.25$

Thus, the probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c)

The probability representing the middle 30% of American men consuming sodium between two weights is:

$P(x_{1}$

Compute the value of <em>z</em> as follows:

$P(-z$

The value of <em>z</em> for P (Z < z) = 0.65 is 0.39.

Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows:

$-z=\frac{x_{1}-\mu}{\sigma}\\-0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8-(0.39\times 0.8)\\x_{1}=9.488\\x_{1}\approx9.5$ $z=\frac{x_{2}-\mu}{\sigma}\\0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8+(0.39\times 0.8)\\x_{1}=10.112\\x_{1}\approx10.1$

Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

4 0

3 years ago