If QN bisects PQR and N is THE midpoint of PR classifiy each triangle by its angles and sides

Five less than the qouicent of number of a number and 3 is -7

Nuetrik [128]

Sorry. Had to redo this with the correct answer.

Here it is...

Solve for x by simplifying both sides of the equation, then isolating the variable.

Exact Form:

x = - 2/5 (fraction)

Decimal Form:

x = - 0.4 (decimal)

7 0

3 years ago

What is the period of f(x) = sin(x)? StartFraction pi Over 2 EndFraction Pi StartFraction 3 pi Over 2 EndFraction 2 pi

Nostrana [21]

<h3>Answer: 2pi</h3>

2pi radians are equal to 360 degrees. After this point, the function f(x) = sin(x) repeats its values again. Each cycle is 2pi radians long.

4 0

3 years ago

Read 2 more answers

What is x - the square root of -2

Lena [83]

Sqrt(-2), can be written as i * sqrt(2). Thus, our answer is x - i sqrt(2).

4 0

3 years ago

3) A school's enrollment in 2019 was 692 students. In 2020 they enrolled 852 students. What was the % increase in students from

elena55 [62]

Answer:

D. 23.1%

Step-by-step explanation:

852-692=160 160/692=0.231213873 rounds to 0.231 move decimal over two places to the right to make it percent which is 23.1%

5 0

3 years ago

A researcher claims that the variation in the salaries of elementary school teachers is greater than the variation in the salari

guajiro [1.7K]

Answer:

a

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

b

$F_{critical} = 1.8608$

c

$F = 2.9085$

d

The decision rule is

Reject the null hypothesis

e

There is sufficient evidence to support the researchers claim

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 30$

The sample variance for elementary school is $s^2_1 = 8324$

The second sample size is $n_2 = 30$

The sample variance for the secondary school is $s^2_2 = 2862$

The significance level is $\alpha = 0.05$

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

Generally from the F statistics table the critical value of $\alpha = 0.05$ at first and second degree of freedom $df_1 = n_1 - 1 = 30 - 1 = 29$ and $df_2 = n_2 - 1 = 30 - 1 = 29$ is

$F_{critical} = 1.8608$

Generally the test statistics is mathematically represented as

$F = \frac{s_1^2 }{s_2^2}$

=> $F = \frac{8324 }{2862}$

=> $F = 2.9085$

Generally from the value obtained we see that $F > F_{critical }$ Hence

The decision rule is

Reject the null hypothesis

The conclusion is

There is sufficient evidence to support the researchers claim

4 0

2 years ago