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4 years ago

Line l has a slope of -7 and a y-intercept of 12. What is the equation for line l in slope intercept form?

Mathematics

1 answer:

djyliett [7]4 years ago

4 0

y = mx + b

is slope intercept form

m is slope

b is y intercept

y = -7x +12

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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

4 0

3 years ago

What is the percent increase modeled in the function f(x)=20(1.5)^x

blagie [28]

The increase is D: 50% because the 1 in the parentheses stands for the 20 that was already there, with the .5 being the amount that’s actually being increased. 0.5 translates to 50%, so the answer is D.

6 0

3 years ago

In a survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for th

marysya [2.9K]

Answer:

The population proportion is estimated to be with 99% confidence within the interval (0.1238, 0.2012).

Step-by-step explanation:

The formula for estimating the population proportion by a confidence interval is given by:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}$

Where:

$\hat{p}$ is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

$z_{\alpha /2}$ is the critical value needed to find the tails of distribution related to the confidence level,

$n$ is the sample's size.

<u>First</u> we compute the $\hat{p}$ value:

$\hat{p}=\frac{successes}{n}=\frac{98}{603}=0.1625$

<u>Next</u> we find the z-score at any z-distribution table or app (in this case i've used StatKey):

$z_{\alpha /2}=2.576$

Now we can replace in the formula with the obtained values to compute the confidence interval:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.1625\pm 2.576\times\sqrt{\frac{0.1625\times(1-0.1625)}{603}}=(0.1238, 0.2012)$