A rectangle has a length x and a width y

Two streets bounding your triangular lot make an angle of 74∘. The lengths of the two sides of the lot on these streets are 126

ycow [4]

Answer: a) Yes, there is enough fance

b) 58.1° and 47.9°

c) The city will not approve, because 1/3 of the area is just 2220.5ft²

Step-by-step explanation:

a) using law of cosines: x is the side we do not know.

x² = 126² + 110² - 2.126.110.cos74°

x² = 20335.3

x = 142.6 ft

So 150 > 142.6, there is enough fance

b) using law of sine:

sin 74/ 142.6 = sinα/126 = sinβ/110

sin 74/ 142.6 = sinα/126

0.006741 = sinα/126

sinα = 0.849

α = sin⁻¹(0.849)

α = 58.1°

sin 74/ 142.6 = sinβ/110

0.006741 = sinβ/110

sinβ = 0.741

β = sin⁻¹(0.741)

β = 47.9°

Checking: 74+58.1+47.9 = 180° ok

c) Using Heron A² = p(p-a)(p-b)(p-c)

p = a+b+c/2

p=126+110+142.6/2

p=189.3

A² = 189.3(189.3-126)(189.3-110)(189.3-142.6)

A = 6661.5 ft²

1/3 A = 2220.5

So 2300 > 2220.5. The area you want to build is bigger than the area available.

The city will not approve

7 0

3 years ago

While at the zoo, you saw 20 monkeys in an exhibit. Then some of the monkeys go and hide while the zoo keeper moves 5 monkeys to

Vladimir79 [104]

It would be 7.

Because 20-5-15 then 15-8=7

7 0

2 years ago

X+(x-10)=26<br> 5,12,15,18

Likurg_2 [28]

Answer: 18

Step-by-step explanation:

put each number in place of x and finish the equation. you will find that 5, 12, and 15 don’t work because they just simply don’t add up.

finally, put 18 in place of x.

18 + (18 - 10) = 26

18 + 8 = 26

26 = 26 ✔️

3 0

2 years ago

Please help me? I am confused.

SIZIF [17.4K]

ANSWER: A. x=4

Explanation: 7-2x=2x(x-4)-1

7-2x=2x-9

-2x-2x=-9-7

-4x=-16

which will get you your answer x=4

7 0

2 years ago

The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) F

MrRissso [65]

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.

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:

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

In this problem, we have that:

$n = 300, p = 0.02$

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

$E(X) = np = 300*0.02 = 6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42$

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

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

$P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040$

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023$

$P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143$

$P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436$

$P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485$

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

From c), we have that $P(X = 0) = 0.0023$ . So

$0.0023 + P(X \geq 1) = 1$

$P(X \geq 1) = 0.9977$

99.77% probability that we find at least 1 of the seniors are married

5 0

3 years ago