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elena55 [62]
3 years ago
12

Which algebraic expression is equivalent to the expression below?

Mathematics
1 answer:
spayn [35]3 years ago
5 0
<span>8(3x + 6) + 13---expand using distributive property
=24x + 48+13 ...combine like terms
=24x +61....simplify

answer is C.24x + 61

hope it helps</span>
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Galina runs a bakery, where she sells packages of 4 dozen cookies for $24.96 per package. The amount of money she makes by selli
elena-s [515]

Answer:

p(x) = f(x) - g(x) = -0.04x² + 20.96x - 71

Explanation:

The sales are given by f(x) = 24.96x and the cost are represented by g(x) = 0.04x² + 4x + 71.

Then, the profit is equal to

p(x) = f(x) - g(x)

p(x) = 24.96x - (0.04x² + 4x + 71)

p(x) = 24.96x - 0.04x² - 4x - 71

p(x) = -0.04x² + 20.96x - 71

Therefore, the answer is

p(x) = f(x) - g(x) = -0.04x² + 20.96x - 71

4 0
2 years ago
Somebody please help me with this
kap26 [50]

Answer:

D. $7/hour

Step-by-step explanation:

128-86=42

42/6=7

6 0
3 years ago
Read 2 more answers
What is the range of f(x)=log0.6x
igor_vitrenko [27]
The range of a log is all real numbers, only the domain is limited. Therefore the answer is +infinity to -infinity.
5 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
4 years ago
We learned that division expressions that have the same quotient and remainder are not necessarily equal to each other explain h
olasank [31]
<span>Let's say you divide 25 by 4.  You will get a quotient 6, but have that remainder of 1.  You could also divide 37 by 6, and likewise get a quotient of 6, with a remainder of 1.  The difference is that the remainders are not truly the same.  The first remainder is 1 part out of 4, but the second remainder is 1 part out of 6</span>
6 0
3 years ago
Read 2 more answers
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