Step 1: –10 + 8x < 6x

Option one(-3, -0.9) and two (-2.5, -0.75) Would not be a viable solution because the value of number of days can not be negative and in option one and two, value of days -3 and -2.5 is negative.

Option three(4.5, 1.35) can not be correct because library charges fee for a full day so the number for days would be a whole number. Library would not charge for 4.5 days, they would either charge of 4 days or 5 days because 4.5 is not an whole number.

Option four(8, 2.40) is the correct answer because it satisfies our equation;

Y= 0.30 * X

2.40= 0.30 * 8

2.40 = 2.40.

Fourth option (8, 2.40) is the only viable solution to this question.

Read more on Brainly.com - brainly.com/question/359512#readmore

8 0

2 years ago

Toyota advertises that their Toyota Highlander Hybrid provide 28 MPG. To test this, an experiment involving a sample of 20 Toyot

klemol [59]

Answer:

We accept the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = 28.8 miles per gallon

Sample size, n = 120

Alpha, α = 0.01

Sample standard deviation, σ = 6.89 miles per gallon

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 28\text{ miles per gallon.}\\H_A: \mu \neq 28\text{ miles per gallon.}$

We use Two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n-1}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{28.8 - 28}{\frac{6.89}{\sqrt{19}} } = 0.5061$

Now, $t_{critical} \text{ at 0.01 level of significance, 19 degree of freedom } =\pm 2.8609$ Since,

$-2.609 < t_{stat} < 2.8609$

We fail to reject the null hypothesis and accept it.

We accept the null hypothesis and the population mean MPG of Toyota Highlander Hybrid vehicles is equal to 28 miles per gallon.

5 0

2 years ago

How many 5-digit numbers are there that are divisible by either 45 or 60 but are not divisible by 90?

o-na [289]

Answer:

2500 numbers

Step-by-step explanation:

Problem;

Solving for the number of 5-digit numbers divisible either by 45 or 60 but not divisible by 90;

Let us approach this in a step wise manner;

5 -digits numbers ranges from 10,000 to 99,999

To solve this problem, let us find the number of digits that are divisible by 45;

the first number divisible by 45 within this range is 10035

the last number divisible by 45 in this range is 99990

increment is 45

The total number in this range is:

$\frac{last number - first number}{increment}$ + 1 = $\frac{99990 - 10035}{45}$ + 1

= 2000 numbers

To solve this problem, let us find the number of digits that are divisible by 60;

the first number divisible by 60 within this range is 10020

the last number divisible by in this range is 99960

increment is 60

$\frac{99990 - 10080 }{90}$

The total number in this range is:

$\frac{last number - first number}{increment}$ + 1 = $\frac{99960 - 10020}{60}$ + 1

= 1500 numbers

To solve this problem, let us find the number of digits that are divisible by 90;

the first number divisible by 90 within this range is 10080

the last number divisible by in this range is 99990

increment is 90

The total number in this range is:

$\frac{last number - first number}{increment}$ + 1 = $\frac{99990 - 10080}{90}$ + 1

= 1000 numbers

Now solution:

Number of 5 digits = number of 45 + number of 60 - number of 90

= 2000 + 1500 - 1000

= 2500 numbers

3 0

2 years ago

Step 1: –10 + 8x < 6x – 4