What is the ratio Between 113 and165

PLease help me with this math. I will give brainly if you get it right!

wel

15 pounds , and about 200 balls

7 0

3 years ago

The endpoints of GE are located at G(–6, –4) and E(4, 8). Using slope-intercept form, write the equation of GE.

lesya692 [45]

Answer:
Equation of the line is:
y = 1.2x + 3.2

Explanation:
The general form of the linear equation is:
y = mx + c
where:
m is the slope
c is the y-intercept

1- getting the slope:
The slope of the line can be calculated using the following formula:
m = $\frac{y2 - y1}{x2 - x1}$

We are given the points:
(-6,-4) representing (x1,y1)
(4,8) representing (x2,y2)
Substitute with the given points in the above equation to get the slope as follows:
m = $\frac{8--4}{4--6} = 6/5$

The equation of the line now is:
y = 1.2x + c

2- getting the y-intercept:
To get the value of the c, we will use any of the given points, substitute in the equation and solve for c.
I will use the point (4,8) as follows:
y = 1.2 x + c
8 = 1.2(4) + c
c = 16/5 = 3.2

Therefore, the equation of the line is:
y = 1.2x + 3.2

Hope this helps :)

4 0

3 years ago

C2 - 6c-5<br> Evaluate the expression when c = -2

vivado [14]

-2(2)-6(-2)-5
-4+12-5
Answer is 3

3 0

3 years ago

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of

kolezko [41]

Answer:

a) Parameter of interest $p$ representing the true proportion of the plates have blistered.

b) Null hypothesis: $p\leq 0.1$

Alternative hypothesis: $p > 0.1$

c) $z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33$

d) For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:

$z_{critc}= 1.28$

e) For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1

f) $p_v =P(z>1.33)=0.0917$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

Part a

Parameter of interest $p$ representing the true proportion of the plates have blistered.

X=14 represent the number of the plates have blistered.

$\hat p=\frac{14}{100}=0.14$ estimated proportion of the plates have blistered.

$p_o=0.1$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

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

Part b: Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that more than 10% of all plates blister under such circumstances.:

Null hypothesis: $p\leq 0.1$

Alternative hypothesis: $p > 0.1$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Part c: Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33$

Part d: Rejection region

For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:

$z_{critc}= 1.28$

Part e

For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1

Part f

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

$p_v =P(z>1.33)=0.0917$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%

7 0

3 years ago

The total attendance at a school play was 1250. The cost of tickets was $6.00 for students

daser333 [38]

Answer:

look at picture

Step-by-step explanation:

x: student tickets

y: adult tickets

7 0

4 years ago