Write an equation of the line passing through point P(4,3) that is parallel to the line y = 1/2(x+4).

An article estimates, with 90% confidence, that the proportion of elementary school teachers who are female is 82%. The article

GREYUIT [131]

Answer: (80% , 85%)

Step-by-step explanation:

We know that, confidence interval for population proportion is given by :-

(p-E , p+E)

, where p = sample proportion , E = Margin of error.

Given : Proportion of elementary school teachers who are female = 82%.

The article also states the maximum error of their estimate = 3%.

Then, the 90% confidence interval for the proportion of elementary school teachers who are female will be :

(82%-2% , 82%+3%)

= (80% , 85%)

Hence, the resulting 90% confidence interval for the proportion of elementary school teachers who are female = (80% , 85%)

5 0

3 years ago

Which is an example of the commutative property?

astra-53 [7]

A. <span>–13 + 27 = 27 + (–13) is an example of the commutative property.

The commutative property states that a + b = b + a.</span>

8 0

3 years ago

if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?

Kay [80]

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the <em>binomial distribution</em> is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

There are 9 rolls, hence $n = 9$ .

Of the six sides, 2 are 3 or 4, hence $p = \frac{2}{6} = 0.3333$

The desired probability is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026$

$P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117$

Then:

$P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857$

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

7 0

3 years ago

2s+4=5(−4−2s)2, s, plus, 4, equals, 5, left parenthesis, minus, 4, minus, 2, s, right parenthesis

Rashid [163]

S= -2 and i need 20 words so ima spam lol sjdjdjfjcjchchf

8 0

3 years ago

30-[11+{18-(48÷8×2)}+7]+15

Inessa05 [86]

30-(11+(18-(48/8×2))+7)+15

= 30-(11+(18-12)+7)+15

= 30-(11+6+7)+15

= 30-24+15

= 21

3 0

3 years ago