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Andrei [34K]
3 years ago
15

P(x) = 4x4 – 9x? – 7x2 – 2x+25

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
8 0

Answer:

27-11x

Step-by-step explanation:

P ( x ) = 4 \times 4 - 9 x ? - 7 \times 2 - 2 x + 25

16-9x-7\times 2-2x+25

16-9x-14-2x+25

2-9x-2x+25

2-11x+25

=27-11x

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Step-by-step explanation:

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The other way you can look at it is that the dimensions of the rectangle are 4 cm and 7 cm. The area of a triangle equation is \frac{bh}{2}. So \frac{4(7)}{2} is 14.

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3 years ago
If f(x) = 5x^2– 3 and f(x + a) = 5x^2+ 30x + 42,<br> what is the value of a ?
Andrei [34K]

Answer:

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Step-by-step explanation:

8 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
What expression is equivilent to -51-(-60)
shutvik [7]

Answer:

-51+60

Step-by-step

When there are two negative or subtraction signs, they form a positive or addition sign.

7 0
3 years ago
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