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bekas [8.4K]
3 years ago
9

(2x+8)(5x-7) A.36x-56 B.10x^2-56 C.10x^2+26x-56 D.10x^2+54x+56

Mathematics
2 answers:
stich3 [128]3 years ago
6 0

<em>so</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>i</em><em>s</em><em> </em><em>of</em><em> </em><em>option</em><em> </em><em>C</em><em>.</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>

valentinak56 [21]3 years ago
3 0

Answer:

The answer is:

C.

10x^2+26x-56

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I think this is right.. I tried my best. If u get it sorry... I’m very sorry



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2.divide 7.20 by 5 ( 5 represents the erasers )

3. Then finally multiply 1.44 from 7.20/5 and
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3 years ago
An amount of $ 8,000 was invested at 6% interest that is compounded yearly. The equation
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8 0
2 years ago
Twelve percent of U.S. residents are in their forties. Consider a group of nine U.S. residents selected at random. Find the prob
miskamm [114]

Answer:

The probability is 28 % ( approx )

Step-by-step explanation:

Since, there are only two possible outcomes in every trails ( residents are in their forties or not ),

So, this is a binomial distribution,

Binomial distribution formula,

Probability of success in x trials,

P(x)=^nC_x p^x q^{n-x}

Where, ^nC_x=\frac{n!}{x!(n-x)!}

p and q are probability of success and failure respectively and n is the total number of trials.

Let X be the event of resident who are in his or her forties.

Here, p = 12% = 0.12,

⇒ q = 1 - p = 1 - 0.12 = 0.88,

n = 9,

Hence, the probability that two or three of the people in the group are in their forties = P(X=2) + P(X=3)

^9C_2 (0.12)^2 (0.88)^{9-2}+^9C_3 (0.12)^3 (0.88)^{9-3}

=36(0.12)^2 (0.88)^7+84(0.12)^3 (0.88)^6

=0.279266611163

\approx 0.28

=28\%

7 0
3 years ago
To win at LOTTO in one​ state, one must correctly select 4 numbers from a collection of 50 numbers​ (1 through 50​). The order i
Oksana_A [137]

Answer:

230,300 different selections.

Step-by-step explanation:

If the order does not matter, the number of possible different selections is determined as the combination of choosing four numbers out of 50:

n=\frac{50!}{(50-4)!4!}\\\\n=\frac{50*49*48*47}{4*3*2*1} \\\\n=230,300\ combinations

There are 230,300 possible different selections.

4 0
3 years ago
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