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Molodets [167]
3 years ago
8

Simplify the problem.

Mathematics
1 answer:
lesya [120]3 years ago
7 0

Answer:

Option 4: 4¹¹

Step-by-step explanation:

Looking at the problem, I need to work out 4⁴ squared first, which is the same as 4⁸. Then multiply that by 4³ to get 4¹¹. What I did was simply add 3 + (4 * 2), which is 11.

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Dmitriy789 [7]
C(1;2)
a = 1
b = 2
r = 4

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(x-1)² + (y-2)² = 4²

(x-1)² + (y-2)² = 16
5 0
4 years ago
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What is the prime factorization of 140? A 2^2×3×7 B. 2^3×7 C. 4×5×7 D. 2^2×5×7​
Andreyy89

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Step-by-step explanation:

7 0
3 years ago
Help please i can’t figure this out
ehidna [41]
To find the y-intercept: replace x with 0
to find the x-intercept: replace y with 0

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6 0
3 years ago
In a​ study, 36​% of adults questioned reported that their health was excellent. A researcher wishes to study the health of peop
steposvetlana [31]

Answer:

0.3907

Step-by-step explanation:

We are given that 36​% of adults questioned reported that their health was excellent.

Probability of good health = 0.36

Among 11 adults randomly selected from this​ area, only 3 reported that their health was excellent.

Now we are supposed to find the probability that when 11 adults are randomly​ selected, 3 or fewer are in excellent health.

i.e. P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)

Formula :P(x=r)=^nC_r p^r q ^ {n-r}

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q = probability of failure = 1- 0.36 = 0.64

n = 11

So, P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)

P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}

P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!}  (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}

P(x\leq 3)=0.390748

Hence  the probability that when 11 adults are randomly​ selected, 3 or fewer are in excellent health is 0.3907

5 0
3 years ago
Which ordered pair is a solution to the equation y = 5x + 2?
Likurg_2 [28]
I’m geussing the answer would be A because an ordered pair of that is (2,8) but the only difference is it’s negative so A hope this helps
6 0
3 years ago
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