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2 years ago

PLEASE HELP ASAP!: Solve for a and b 6a-b=-5 4a-3b = -8

Mathematics

1 answer:

klemol [59]2 years ago

7 0

Answer:

a = -1/2

b = 2

Step-by-step explanation:

Step 1: Rewrite 1st equation

-b = -5 - 6a

b = 5 + 6a

Step 2: Substitution

4a - 3(5 + 6a) = -8

Step 3: Solve

4a - 15 - 18a = -8

-14a - 15 = -8

-14a = 7

a = -1/2

Step 4: Plug in a to find b

6(-1/2) - b = -5

-3 - b = -5

-b = -2

b = 2

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Eva8 [605]

Answer:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=250000$ represent the sample mean

$s=37500$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =210250$ represent the value that we want to test

$\alpha=0.1$ 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 mean is higher than 210250, the system of hypothesis would be:

Null hypothesis: $\mu \leq 210250$

Alternative hypothesis: $\mu > 210250$

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:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

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

$df=n-1=5-1=4$

Conclusion

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

$p_v =P(t_{4}>2.37)=0.038$

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Write an equation that goes through (8,1) and is perpendicular to 2y + 4x =12

Serggg [28]

Answer:

2y -x + 6 = 0

Step-by-step explanation:

Here a equation and a point is given to us and we are interested in finding a equation which is perpendicular to the given equation and passes through the given point .

The given equation is ,

$\sf \longrightarrow 2y + 4x = 12 \\$

$\sf \longrightarrow 4x + 2y = 12$

Firstly convert this into slope intercept form , to find out the slope of the line.

$\sf \longrightarrow 2y = -4x +12\\$

$\sf \longrightarrow y =\dfrac{-4x+12}{2}\\$

$\sf \longrightarrow y =\dfrac{-4x}{2}+\dfrac{12}{2}\\$

$\sf \longrightarrow \red{ y = -2x + 6 }$

Now on comparing it to slope intercept form which is y = mx + c , we have ,

m = -2

And as we know that the product of slopes of two perpendicular lines is -1 . So the slope of the perpendicular line will be negative reciprocal of the slope of the given line. Therefore ,

$\sf \longrightarrow m_{\perp}=\dfrac{-1}{-2}=\red{\dfrac{1}{2}}$

Now we may use the point slope form of the line to find out the equation of the line using the given point . The point slope form is,

$\sf \longrightarrow y - y_1 = m(x - x_1)$

Now on substituting the respective values we have,

$\sf \longrightarrow y - 1 = \dfrac{1}{2}(x-8)\\$

$\sf \longrightarrow 2(y-1 )= x -8 \\$

$\sf \longrightarrow 2y -2=x-8\\$

$\sf \longrightarrow \underline{\boxed{\bf 2y - x + 6=0}}$

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2 years ago

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2 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

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Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ subtract 16 from both sides

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$

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3 years ago