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2 years ago

Find 116% of 75. please i need to show my work so please help with the work WITH THE WORK SHOWN

Mathematics

2 answers:

snow_lady [41]2 years ago

8 0

Answer: 87

Step-by-step explanation:

116% = * 1.16

75 * 1.16 = 87

Andre45 [30]2 years ago

6 0

Answer: 87

Step-by-step explanation:

75 percent *116 =

(75:100)*116 =

(75*116):100 =

8700/100 = 87

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Suppose a triangle has two sides of length 3and 4 and that the angle between these two sides is 60. What is the length of the th

tangare [24]

9514 1404 393

Answer:

A. √13

Step-by-step explanation:

You can make an educated guess and come to the right conclusion.

The triangle is nearly an equilateral triangle. A triangle with two sides 3 and an angle of 60° would have a third side of 3. A triangle with two sides of 4 and an angle of 60° would have a third side of 4.

So, the third side must be between 3 and 4. Here is an evaluation of the answer choices:

A -- between 3 and 4, the correct choice

B -- 3, too short

C -- 1.73, too short

D -- more than 4, too long

The question can be answered using your triangle solver app on your calculator, or using the Law of Cosines.

c = √(a^2 +b^2 -2ab·cos(C))

c = √(3^2 +4^2 -2·3·4·(1/2)) = √(9 +16 -12)

c = √13 . . . . . length of the side opposite the 60° angle

5 0

3 years ago

Psychologist Michael Cunningham conducted a survey of university women to see whether, upon graduation, they would prefer to mar

bagirrra123 [75]

Answer:

A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.

B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.

C. The expected value of X is 6.75, and the standard deviation of X is 2.17.

Step-by-step explanation:

The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\$

A. P(x=6)

$P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\$

B. P(x≥10)

$P(x\geq10)=1-P(x$

$P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\$

$P(x=7) = \dbinom{25}{7} p^{7}(1-p)^{18}=480700*0.000061*0.005638=0.1654\\\\\\P(x=8) = \dbinom{25}{8} p^{8}(1-p)^{17}=1081575*0.000015*0.007517=0.1241\\\\\\P(x=9) = \dbinom{25}{9} p^{9}(1-p)^{16}=2042975*0.000004*0.010023=0.0781\\\\\\$