What is the inverse of y=1/4X-4

Mrs. Simmons gave a history test worth 92 points. There were only two types of questions on it: 2-point true/false questions and

SVETLANKA909090 [29]

Answer:

There are 8 true/false questions and 26 fill-in-the-blank questions.

Step-by-step explanation:

Let "t" be the number of 2-point true/false questions and "f" the number of 5-point fill-in-the-blank questions.

Mrs. Simmons gave a history test worth 92 points. Symbollically,

2 t + 5 f = 92 [1]

There were a total of 34 questions in the test. Symbollicaly,

t + f = 34

t = 34 - f [2]

If we replace [2] in 1, we get

2 (34 - f) + 5 f = 92

68 - 2 f + 5 f = 92

3 f = 24

f = 8

We replace f = 8 in [2],

t = 34 - 8 = 26

There are 8 true/false questions and 26 fill-in-the-blank questions.

7 0

3 years ago

A number cube with the numbers 1-6 is rolled 1000 times about how many times would it be expected that a number less than 5 is r

mamaluj [8]

666 times is what it should be

7 0

3 years ago

Read 2 more answers

What is the length of the diagonal, d, of the rectangular prism shown below?

melisa1 [442]

The answer is 7 units

6 0

3 years ago

Speed limit signs are placed every 58 mi on the highway. How many signs are there on a 75 mi stretch of highway? Which operation

ki77a [65]

Answer:

you should use multiatication

Step-by-step explanation:

5 0

3 years ago

Consider the hypotheses below. Upper H 0: mu equals 50 Upper H 1: mu not equals 50 Given that x overbar equals 60, s equals 1

Neko [114]

Answer:

We conclude that $\mu\neq$ 50.

Step-by-step explanation:

We are given that x bar = 60, s = 12, n = 30, and alpha = 0.10

Also, Null Hypothesis, $H_0$ : $\mu$ = 50

Alternate Hypothesis, $H_1$ : $\mu$ $\neq$ 50

The test statistics we will use here is;

T.S. = $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, X bar = sample mean = 60

s = sample standard deviation = 12

n = sample size = 30

So, Test statistics = $\frac{60-50}{\frac{12}{\sqrt{30} } }$ ~ $t_2_9$

= 4.564

At 10% significance level, t table gives critical value of 1.311 at 29 degree of freedom. Since our test statistics is more than the critical value as 4.564 > 1.311 so we have sufficient evidence to reject null hypothesis as our test statistics will fall in the rejection region.

P-value is given by, P( $t_2_9$ > 4.564) = less than 0.05% {using t table}

Here also, P-value is less than the significance level as 0.05% < 10% , so we will reject null hypothesis.

Therefore, we conclude that $\mu\neq$ 50.

7 0

3 years ago