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Tanya [424]
3 years ago
13

Question 13

Mathematics
1 answer:
kicyunya [14]3 years ago
4 0

Answer:

The c intercept is 42

The t intercepts are: 6, -1 and 7

Step-by-step explanation:

Given

c(t) = (t - 6)(t +1)(t-7)

Solving (a): The c intercept

Simply set t to 0

c(t) = (t - 6)(t +1)(t-7)

c(0) = (0 - 6)(0 +1)(0-7)

c(0) = (- 6)(1)(-7)

c(0) = 42

Solving (b): The t intercept

Simply set c(t) to 0

c(t) = (t - 6)(t +1)(t-7)

(t - 6)(t +1)(t-7) = 0

Split

t - 6= 0,\ \ t +1= 0,\ \ t-7 = 0

Solve for t

t = 6,\ \ t =-1,\ \ t=7

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The Wilson family had 5 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Wi
Dafna11 [192]

Answer:

0.813

0.500

Step-by-step explanation:

Use binomial probability.

P = nCr p^r q^(n−r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

In this problem, n = 5, p = 0.5, and q = 0.5.

"At least 2 girls" means r = 2, 3, 4, or 5.

Or, we can use the complement.

P(at least 2 girls) = 1 − P(at most 1 girl)

P(at least 2 girls) = 1 − P(r=0 or r=1)

P(at least 2 girls) = 1 − ₅C₁ (0.5)¹ (0.5)⁵⁻¹ − ₅C₀ (0.5)⁰ (0.5)⁵⁻⁰

P(at least 2 girls) = 1 − 5 (0.5) (0.5)⁴ − 1 (1) (0.5)⁵

P(at least 2 girls) = 1 − 6 (0.5)⁵

P(at least 2 girls) ≈ 0.813

"At most 2 girls" means r = 0, 1, or 2.

P(at most 2 girls) = P(r=0, r=1, or r=2)

P(at most 2 girls) = ₅C₀ (0.5)⁰ (0.5)⁵⁻⁰ + ₅C₁ (0.5)¹ (0.5)⁵⁻¹ + ₅C₂ (0.5)² (0.5)⁵⁻²

P(at most 2 girls) = 1 (1) (0.5)⁵ + 5 (0.5) (0.5)⁴ + 10 (0.5)² (0.5)³

P(at most 2 girls) = 16 (0.5)⁵

P(at most 2 girls) = 0.500

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tell us the question so we can help.

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9x - y = 15 2x + 8y = 28
faust18 [17]
9x - y = 15
2x + 8y = 28

Use the substitution method.
Solve for y in the first equation.

9x - y = 15
-y = 15 - 9x
y = -15 + 9x

Now plug in y into the second equation.
2x + 8(-15 + 9x) = 28
2x - 120 + 72x = 28
74x - 120 = 28
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Plug x back into the rewritten first equation.
y = -15 + 9(2)
y = -15 + 18
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x = 2, y = 3
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