The area A of a rectangle with a width of x-1 is given by A=2x^4+x

Write the number 3.8 in the form a/b using integers

Juliette [100K]

Answer:

3.8 = 38/10

Step-by-step explanation:

The given number = 3.8

We have to write 3.8 in fraction form a/b

Here we have 1 digit after the decimal. To get rid off the decimal number, we need to multiply both the numerator and the denominator by 10 since we have to 1 digit after the decimal.

$\frac{3.8 *10}{10}$

= 38/10

Therefore, the answer is 3.8 = 38/10

Thank you.

6 0

4 years ago

You lose 3 pounds per week on your new diet. What integer represents the change in your weight after 3 weeks

horsena [70]

Hello,

If you lose 3 pounds per week, after 3 weeks you lose (3)(3) = 9 pounds.

The change in your weight is represented by the integer 9 (pounds).

Hope this helps!

3 0

3 years ago

Read 2 more answers

15- pls help!!!! math pls help answer other questions too

ddd [48]

Answer:

A

Step-by-step explanation:

The projection of vector B on unit vector a is

(B•a)a

av(unit vector of v)= 1/√2(-1,1)

u•av= 1/√2 × (-5+7)= √2

(u•av)av= √2 × 1/√2(-1,1)= (-1,1)

3 0

3 years ago

The table below represents a linear relationship.

Ede4ka [16]

Answer:i think it 3

Step-by-step explanation:

5 0

3 years ago

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re

Zolol [24]

Answer:

a) $df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

b) $df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

c) $df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =20$ represent the value that we want to test

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

t 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 determine if the true mean is higher than 20, the system of hypothesis would be:

Null hypothesis: $\mu \leq 20$

Alternative hypothesis: $\mu > 20$

We don't know the population deviation, so for this case 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".

(a) n = 16, t = 3.3, a = 0.05, P-value =

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

$df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

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

$df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

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

$df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

3 0

4 years ago

The area A of a rectangle with a width of x-1 is given by A=2x^4+x - 7x + 4. Find an expression for the