Helppppppppppp please...

Can someone help me with these problems? Please!

masya89 [10]

Step-by-step explanation:

2) Slope can be written as rise over run:

rise: add to the y-value
run: add to the x-value

$a) Slope=-1\ over\ 1\quad \rightarrow \quad (11 + 1, -8 - 1)\implies (12, -9)\\b) Slope=7\ over\ 2\quad \rightarrow \quad (2+2,2+7)\implies (4, 9)\\c) Slope=9\ over\ 4\quad \rightarrow \quad (-5+4,-2+9)\implies (-1, 7)$

3) Slope formula is: $\dfrac{y_2-y_1}{x_2-x_1}$

$slope=\dfrac{15-3}{5-1}=\dfrac{12}{4}=\boxed{3}$

4) Average rate of change is the slope of the given interval:

a) when x = 0, y = 0 and when x = 1, y = -3

$Slope=\dfrac{-3-0}{1-0}=\dfrac{-3}{1}=-3$

b) when x = -2, y = 0 and when x = -1, y = 3

$Slope=\dfrac{3-0}{-1+2}=\dfrac{3}{1}=3$

c) when x = 0, y = 0 and when x = 2, y = 0

$Slope=\dfrac{0-0}{2-0}=\dfrac{0}{2}=0$

3 0

4 years ago

the graph of a liner function passes through the two gives ponits on the coordintes plne (3,15) (6,21) what is the rate of chang

slega [8]

Answer:

The rate of change of the function is 2.

Step-by-step explanation:

As we know that if the graph of a liner function passes through the two gives points on the coordinate plane, then the rate of change of the function will be the slope of the given line.

The points are:

(3, 15)

(6, 21)

$\mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(3,\:15\right),\:\left(x_2,\:y_2\right)=\left(6,\:21\right)$

$m=\frac{21-15}{6-3}$

$m=2$

Therefore, the rate of change of the function is 2.

8 0

3 years ago

Fountain drink sizes StartSet small comma medium comma large EndSet Are the data qualitative or quantitative?

klemol [59]

Answer:

Qualitative

Step-by-step explanation:

Qualitative describes sizes of drinks in comparison to each other.
Quantitative sizes should include measurements in a standarized system (e.g. the imperial system, or the metric system).

4 0

3 years ago

How these question will get solve ? Please help.

nlexa [21]

Ok so look

we have
2√3+3√3-4√3
easy
remember taht 2√3 means 2 times √3 or √3+√3
it tells how many of that
so just add the numbers
2+3-4=1
1√3 or √3 is answer

2.
first simplfy each radical

5√32=5(√16)(√2)=5(4)(√2)=20√2
2√50=2(√25)(√2)=2(5)(√2)=10√2
now we have
20√2+10√2
just add
30√2

1. √3
2. 30√2

4 0

3 years ago

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

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

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ 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 to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

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".

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

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

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

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

3 0

3 years ago