All categories

3 years ago

Please helps!!!!!!!!!!!!!!

Mathematics

1 answer:

Goshia [24]3 years ago

3 0

Statement Reason

---------------------------------------------

1. RS ≅ UT | Given

2. RT ≅ US | Given

3. ST ≅ TS | Reflexive Property

4. △RST≅△UTS | SSS

You might be interested in

James is 5 years younger than his sister. The sum of their ages is 25. Write a system of equation to model the situation. How ol

Sliva [168]

Answer:15

Step-by-step explanation:So Adam is 20 years old, and Neil is 5 years younger so is 20 - 5 = 15 years old. The ratio of Adam's age to Neil's age is 20:15, or 4:3.

4 0

4 years ago

Suppose that the times required for a cable company to fix cable problems in its customers’ homes are uniformly distributed betw

vampirchik [111]

Answer:

a) 60% probability that a randomly selected cable repair visit will take at least 50 minutes

b) 60% probability that a randomly selected cable repair visit will take at most 55 minutes.

c) The expected length of the repair visit is 52.5 minutes.

d) The standard deviation is 7.22 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

For this problem, we have that:

Uniformly distributed between 40 and 65 minutes, so $a = 40, b = 65$

(a) What is the probability that a randomly selected cable repair visit will take at least 50 minutes?

Either it takes less than 50 minutes, or it takes at least 50 minutes. The sum of the probabilities of these events is decimal 1. So

$P(X < 50) + P(X \geq 50) = 1$

The intervals can be open or closed, so

$P(X < 50) = P(X \leq 50)$

We have that

$P(X \geq 50) = 1 - P(X < 50)$

$P(X < x) = \frac{50 - 40}{65 - 40} = 0.4$

$P(X \geq 50) = 1 - P(X < 50) = 1 - 0.4 = 0.6$

60% probability that a randomly selected cable repair visit will take at least 50 minutes

(b) What is the probability that a randomly selected cable repair visit will take at most 55 minutes?

This is $P(X \leq 55)$

$P(X \leq 55) = \frac{55 - 40}{65 - 40} = 0.6$

60% probability that a randomly selected cable repair visit will take at most 55 minutes.

(c) What is the expected length of the repair visit?

The mean of the uniform distribution is:

$M = \frac{a+b}{2}$

$M = \frac{40+65}{2} = 52.5$

The expected length of the repair visit is 52.5 minutes.

(d) What is the standard deviation?

The standard deviation of the uniform distribution is

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

$S = \sqrt{\frac{(65-40)^{2}}{12}}$

$S = 7.22$

The standard deviation is 7.22 minutes.

5 0

3 years ago

Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0

Bas_tet [7]

Sin 2x - sin 4x = 0
sin 2x - 2sin 2x cos 2x = 0
sin x(1 - 2cos 2x) = 0
sin x = 0 or 1 - 2cos 2x = 0
sin x = 0 or 2cos 2x = 1
sin x = 0 or cos 2x = 1/2
x = arc sin 0 or 2x = arc cos (1/2)
x = arc sin 0 or x = 1/2 arc cos (1/2)
x = 0, π, 2π or x = π/6, 7π/6

3 0

3 years ago

Read 2 more answers

1. The national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15. The dean of a college wants to kno

Nat2105 [25]

Answer:

We conclude that the mean IQ of her students is different from the national average.

Step-by-step explanation:

We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.

The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113

Let, Null Hypothesis, $H_0$ : $\mu$ = 100 {means that the mean IQ of her students is same as of national average}

Alternate Hypothesis, $H_1$ : $\mu\neq$ 100 {means that the mean IQ of her students is different from the national average}

(a) The test statistics that will be used here is One sample z-test statistics;

T.S. = $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }$ ~ N(0,1)

where, Xbar = sample mean score = 113

s = population standard deviation = 15

n = sample of students = 144

So, test statistics = $\frac{113-100}{\frac{15}{\sqrt{144} } }$

= 10.4

Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean IQ of her students is different from the national average.

3 0

4 years ago

Show do i solve for the missing answers

Maurinko [17]

(g + f)(x)
g(x) + f(x)
x - 1 + x^2 + 4
x^2 + x + 3

(g - h)(x)
g(x) - h(x)
3x + 1 - (2x - 2)
3x + 1 - 2x + 2
x + 3

8 0

2 years ago