Ask question

Ask question

All categories

3 years ago

6

Marcie wants to enclose her yard with a fence. Her yard is in the shape of a rectangle attached to a triangle. The formula for t

he area of the enclosed space is A = lw + 0.5bh. Given the formula, find the value of b. A. b = A divided by the quantity l times w plus 0.5 times h B. b = The quantity A minus l times w all divided by 0.5 times h C. b = A – lw – 0.5h. . D. b = A + lw + 0.5h

1 answer:

Lana71 [14]3 years ago

3 0

A = lw + 0.5bh
0.5bh = A - lw
b = (A - lw)/0.5h

option B is the correct option.

You might be interested in

62% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

dedylja [7]

Answer:

a) 0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b) 0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c) 0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d) 0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

Step-by-step explanation:

To solve this question, we use the binomial probability distribution, and also it's approximation to the normal distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

62% of owned dogs in the United States are spayed or neutered.

This means that $p = 0.62$

48 owned dogs are randomly selected

This means that $n = 48$

Mean and standard deviation, for the approximation:

$\mu = E(x) = np = 48*0.62 = 29.76$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{48*0.62*0.38} = 3.36$

a. Exactly 30 of them are spayed or neutered.

This is P(X = 30), which is not necessary the use of the approximation.

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

$P(X = 30) = C_{48,30}.(0.62)^{30}.(0.38)^{18} = 0.1180$

0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b. At most 33 of them are spayed or neutered.

Now we use the approximation. This is, using continuity correction, $P(X \leq 33 + 0.5) = P(X \leq 33.5)$ , which is the pvalue of Z when X = 33.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{33.5 - 29.76}{3.36}$

$Z = 1.11$

$Z = 1.11$ has a pvalue of 0.8665

0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c. At least 31 of them are spayed or neutered.

Using continuity correction, this is $P(X \geq 31 - 0.5) = P(X \geq 30.5)$ , which is 1 subtracted by the pvalue of Z when X = 30.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30.5 - 29.76}{3.36}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

1 - 0.5871 = 0.4129

0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d. Between 24 and 30 (including 24 and 30) of them are spayed or neutered.

This is, using continuity correction, $P(24 - 0.5 \leq X \leq 30 + 0.5) = P(23.5 \leq X \leq 30.5)$ , which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 23.5.

X = 30.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30.5 - 29.76}{3.36}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

X = 23.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{23.5 - 29.76}{3.36}$

$Z = -1.86$

$Z = -1.86$ has a pvalue of 0.0314

0.5871 - 0.0314 = 0.5557

0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

8 0

3 years ago

Equation, find y when x=-3.<br><br> y = -x + 5<br><br> When x = -3, y =

Step2247 [10]

I think y = 8. Since x = -3 then the equation would be y = -1(-3) +5. -1 times -3 equals 3. So now the equation is y = 3 +5. That would give you 8.

7 0

3 years ago

Multiply. <br><br> 4 × 0.002 =

trapecia [35]

The answer to your question is 0.008

7 0

3 years ago

2x+2(1960/x)=200 please help for math hw

sp2606 [1]

Two solutions:

1. x =(100-√2160)/2=50-6√ 15 = 26.762

2. x =(100+√2160)/2=50+6√ 15 = 73.238

3 0

3 years ago

Please find the missing side of this triangle

Strike441 [17]

The answer is 15.01

First you need to find the other angle of the triangle so you do 90+25 which gives you 115 and subtract that by 180 because all angles of a triangle add up to 180. That gives you 65 then you will use TAN(65) and get 2.144506920509559 which you will then multiple by 7 to get 15.01154844356691 which you then round to get 15.01!

7 0

3 years ago