Find the percent change from$20 to $25.

The graph of the function has slope of 1 and y-intercept of -6.

True [87]

Answer:

= -5

Step-by-step explanation:

1-6=5

1-6

{Subtract the numbers:} 1-6=-5

= -5

6 0

2 years ago

Prove: An odd number squared is an odd number. = (2n + 1)2

Mnenie [13.5K]

Answer:

${(2n + 1)}^{2} = 4 {n}^{2} + 4n + 1 = 2(2 {n}^{2} + 2n) + 1$

4 0

2 years ago

What are all the values of x that make the quadratic A<br> equation below true?<br> x² – 16=0

Rudiy27

X^2 = 16
x = 4
x = -4
good luck :)

8 0

2 years ago

The sum of two numbers is less than 2. If we subtract the second number from the first, the difference is greater than 1. What a

Anna007 [38]

Answer:

the value of these two numbers are 1/2 and 3/2

Step-by-step explanation:

Given that:

x+y < 2 ---- (1)

y - x > 1 ---- (2)

From equation (2), let y > 1 + x then substitute it into equation (1)

x + 1 + x < 2

2x + 1 < 2

2x < 2 - 1

2x < 1

x < 1/2

From equation (2), replace the value of x to be 1/2

y - 1/2 > 1

y > 1 + 1/2

y > 3/2

5 0

2 years ago

Some research suggests that police officers are more likely to make an arrest in the presence of bystanders. If mentally disorde

zmey [24]

Answer:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

$p_v =P(t_{(19)}>1.638)=0.0589$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=7.03$ represent the sample mean

$s=9.42$ represent the sample standard deviation

$n=20$ sample size

$\mu_o =3.58$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 3.58 :

Null hypothesis: $\mu \leq 3.58$

Alternative hypothesis: $\mu > 3.58$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since is a right tailed test the p value would be:

$p_v =P(t_{(19)}>1.638)=0.0589$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

7 0

3 years ago