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3 years ago

Please help What is the slope of a line that is parallel to the line shown?

Mathematics

2 answers:

9966 [12]3 years ago

4 0

4/6 and (0,1) is the answers

zavuch27 [327]3 years ago

4 0

Answer:

slope of parallel line= 2/3

Step-by-step explanation:

1. Find equation of line

a). y-intercept=1

b). slope= rise over run from one point to another

c). slope= 2/3

d). plug in equation to y=mx+b form

e).equation= y=2/3x+1

2. Find parallel line

a). two parallel lines have the same slope and a different y-intercept

Answer= the same slope as the original line (aka. 2/3)

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48 logs on 6 trucks

blagie [28]

Answer:

288 is your answer. 8 is your answer.

Step-by-step explanation:

If you want to know how many logs all together, all you do is multiply.

48 x 6 = 288

288 is your answer.

Or division it is

48/6 = 8

8 is your answer.

6 0

3 years ago

If a quadratic equation with real coefficients has a discriminant of 3 then the two roots must be

lukranit [14]

In a quadratic equation

q(x) = ax^2 + bx + c

The discriminant is = b^2 - 4ac

We have that discriminant = 3

If b^2 - 4ac > 0, then the roots are real.

If b^2 - 4ac < 0 then the roots are imaginary

<span>In this problem b^2 - 4ac > 0 3 > 0 </span>

then the two roots must be real

6 0

3 years ago

If 8% of all cell phones are defective, and you have a random sample group of 100 Samsung Galaxy Vills,

Sati [7]

Answer:

Mean = 8

Variance = 7.36

Standard Deviation = 2.7129

Step-by-step explanation:

This is a binomial distribution with parameters, n and p.

Where

n is sample size (given as 100)

p is the probability of success, or probability of defective (given as 8% or 0.08)

The mean, variance, standard deviation formula for binomial distribution is shown below:

Mean = $np$

Variance = $npq$

Standard Deviation = $\sqrt{npq}$

Where q would be probability of failure, or "1 - p"

Thus,

n = 100

p = 0.08

q = 1 - 0.08 = 0.92

SO, we have:

Mean = $np=100*0.08=8$

Variance = $npq=100*0.08*0.92 = 7.36$

Standard Deviation = $\sqrt{npq}=\sqrt{7.36}=2.7129$

5 0

3 years ago

4.35 divided by 5? does anyone know??????

Dvinal [7]

.87 is correct you can always check your work with an online calculator. :)

5 0

3 years ago

A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading

valkas [14]

Answer:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

$p_v =2*P(t_{11}$

Step-by-step explanation:

1) Data given and notation

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

The results obtained are:

$\bar X=98.375$ represent the sample mean

$s=0.6.109$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =100$ represent the value that we want to test

$\alpha$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

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".

3) Calculate the statistic

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

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

4) P-value

First we need to find the degrees of freedom for the statistic given by:

$df=n-1=12-1=11$

Since is a two sided test the p value would given by:

$p_v =2*P(t_{11}$

5) Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.

3 0

3 years ago