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3 years ago

Evaluate. Please answer! I'd appreciate it! ASAP

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

5 0

$3 \sqrt{7} \sqrt{7} = 3 \sqrt{ {7}^{2} } = 3 \times 7 = 21 \\ 6 \sqrt{ {7}^{5} } = 6 \sqrt{ {7}^{4} \times 7 } = 6 \times {7}^{2} \sqrt{7} = 294 \sqrt{7} \\ \frac{21}{294 \sqrt{7} } = \frac{1}{14 \sqrt{7} } \\ \\ or \: \frac{1}{14 \sqrt{7} } \times \frac{ \sqrt{7} }{ \sqrt{7} } = \frac{ \sqrt{7} }{14 \times 7} = \frac{ \sqrt{7} }{98}$

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ser-zykov [4K]

Raise each number to the power outside the parentheses.

$\left( \dfrac{81x^{12}}{3x^3} \right) ^{ \frac{1}{3} } = \\

= \left( 27x^{12-3} \right) ^{ \frac{1}{3} } = \\

= \left( 27x^{9} \right) ^{ \frac{1}{3} } = \\

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= \left( 3^3 \right) ^{ \frac{1}{3} } \times \left( x^{9} \right) ^{ \frac{1}{3} } \\

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3 years ago

How do you solve c for a/b=c

asambeis [7]

The variable your solving for should be on the left side, so just flip it
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3 years ago

According to the Center for Disease Control and Prevention (CDC), up to 20% of Americans contract the influenza virus each year,

12345 [234]

Answer:

(1) a. 0.0009

(2) d. 0.640

(3)

a. P(A and B) = 0.06.
b. P(A or B) = 0.70.

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(5) a. nearly 0.

(6)b. 0.919

Step-by-Step Explanation:

(1)Probability of a baby being born with a birth defect =3%=0.03

The probability that both babies have birth defects=0.03 X 0.03= 0.0009.

(2)The probability of contracting the influenza virus each year = 20%=0.2

Therefore, the probability of not contracting the influenza virus =1-0.2=0.8

The probability that neither baby catches the flu in a given year:

=0.8 X 0.8

=0.64

(3)

P(A)=0.1

P(B)=0.6

P(A or B)=P(A)+P(B)=0.1 + 0.6 =0.7

P(A and B)=P(A)XP(B)=0.1 X 0.6 =0.06

(4)

P(A)=0.2

P(B)=0.9

Event A and B cannot be disjoint.

(5)

The probability of an American woman aged 20 to 24 having Chlamydia infection $=\dfrac{2791.5}{100000}$

The probability that three randomly selected women in this age group have the infection

$=\dfrac{2791.5}{100000} \times \dfrac{2791.5}{100000} \times \dfrac{2791.5}{100000} \\\\=0.00002175\\\approx 0$

(6)The probability of an American woman aged 20 to 24 not having Chlamydia infection $=1-\dfrac{2791.5}{100000}$

The probability that three randomly selected women in this age group do not have the infection

$=\left(1-\dfrac{2791.5}{100000}\right)^3\\\\=0.9186\\\approx 0.919$

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4 years ago

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ratelena [41]

Compute the volume of $Q$ :

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So the average value of $f$ over $Q$ is 2/(4/3) = 3/2.

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4 years ago

Please help , giving brainliest and a thanks!!

pashok25 [27]

Answer:

Step-by-step explanation:

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3 years ago

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