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borishaifa [10]
3 years ago
11

Jed's has $45 for the amusement park. He spent $17 for admission into the park.

Mathematics
1 answer:
stellarik [79]3 years ago
4 0
45-17=28
the answer is 28
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I believe it’s 7/10
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c

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Do 28.5/42 and 19/28 form a proportion​
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Yes they do!

Step-by-step explanation:

Both fractions simplify into approximately 0.68

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What is the conjugate of -3+2i
WARRIOR [948]
\text{conjugate:\ \boxed{ -3-2i}}
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Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask
Kay [80]

Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not  feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

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

Where, ^nC_r=\frac{n!}{r!(n-r)!}

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}

=56(0.4)^5 (0.6)^3

=0.12386304

(b) the probability of the number that say they would feel secure is more than five

P(X>5) = P(X=6)+ P(X=7) + P(X=8)

=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}

=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8

=0.04980736

(c) the probability of the number that say they would feel secure is at most five

P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}

=0.6^8+8(0.4)(0.6)^7+28(0.4)^2(0.6)^6+56(0.4)^3(0.6)^5+70(0.4)^4(0.6)^4+56(0.4)^5(0.6)^3

=0.95019264

8 0
3 years ago
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