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borishaifa [10]
3 years ago
7

Evaluate f(x) = 4•5 for x=1

Mathematics
1 answer:
sweet [91]3 years ago
7 0

Answer:

Not possible

Step-by-step explanation:

Unless you may have entered the question wrong, that cannot work

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-1/2(-3y+10) i need help combining like terms to create a equivelent expression
Vaselesa [24]

Answer:

1.5y - 5

Step-by-step explanation:

I used the distributive approach and distributed -1/2 to the -3y and 10.

(If you try this and it's incorrect, you may just need to distribute -1/2 to the 10, creating -3y - 5.)

6 0
3 years ago
-4(2x-3) + 4x = 9x -4x +8 -4 answer this
shtirl [24]

Answer:

x = 8/9

Step-by-step explanation:

-4(2x-3) + 4x = 9x -4x +8 -4  

2x*(-4) -3*(-4) + 4x = 9x - 4x  + 8 - 4

-8x + 12 + 4x = 9x - 4x + 8 - 4         {Combine like terms}

-4x + 12 = 5x + 4

12  = 5x + 4x + 4

12 = 9x + 4

12- 4 = 9x

9x = 8

x = 8/9

5 0
3 years ago
Help me please i need it
crimeas [40]

Answer:

no picture

Step-by-step explanation:

7 0
3 years ago
Max is making a rectangular garden that is 5 feet less than twice it’s width. If the perimeter of the garden is 80 feet what wil
mixas84 [53]

Answer:

width is 15 feet and length is 25 feet

(brainly.com/question/3733624)

6 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7
Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

    The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable <em>X</em> ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 2.7 and standard deviation, <em>σ</em> = 0.75.

(a)

A random sample of <em>n</em> = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

P(\bar X\leq 3.00)=P(\bar X

                    =P(\bar X

                    =P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

                    =P(Z

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

P(2.70

                                =P(0

*Use a <em>z</em>-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

P(\bar X\leq 3.00)\geq 0.99

P(\bar X

P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

P(Z

The value of <em>z</em> for the probability above is

<em>z</em> ≥ 2.33

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

\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33

\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}

\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77

Thus, the sample size must be at least 77.

8 0
3 years ago
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