The quotient of the sum of r and s spilt into 3 equal groups

Which inequality will have a shaded area above its graph?

pochemuha

x − 9y ≤ 1

Because If you solve the equation y will be greater than ...

x − 9y ≤ 1

Solve for y...

Subtract x from both sides...

- 9y ≤ - x + 1

Divide both sides by -9

REMEMBER WHEN DIVIDING OR MULTIPLYING BY A NEGATIVE NUMBER FLIP THE INEQUALITY SIGN....

y ≥ x/9 - 1/9

7 0

3 years ago

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The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

C

b

C

c

C

d

A

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

PLEASE ANSWER I WILL GIVE BRAINLIEST

Naddik [55]

Should be the first option.

Since ABP is congruent to CBP in the 4th step, AP must be equal to CP

6 0

4 years ago

Help please (i pressed 0.09 by mistake)

Alja [10]

Answer:

The answer should be .25

Step-by-step explanation:

The answer should be .25 because:

- there are 4 places you can score and since you want it to land on 5 and there is only one 5 it counts as a 1

- the since there are 4 in total and you want to land on 1 you get 1/4 the number one standing for there only being one 5 and the 4 standing for the number of places to land on.

-then you convert the 1/4 into a decimal which makes .25

4 0

3 years ago

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1:Under what condition will the line px+py+r=0 mat be a normal to the circke x²+y²+2gx+2fy+c=0

ahrayia [7]

Answer:

The normal overlaps with the diameter, so it passes through the center.

Let's find the center of the circle:

x² + y² + 2gx + 2fy + c = 0
(x + g)² + (y + f)² = c + g² + f²

The center is:

(-g, -f)

Since the line passes through (-g, -f) the equation of the line becomes:

p(-g) + p(-f) + r = 0
r = p(g + f)

This is the required condition

Rewrite equations and find centers and radius of both circles.

Circle 1

x² + y² + 2ax + c² = 0
(x + a)² + y² = a² - c²
The center is (-a, 0) and radius is √(a² - c²)

Circle 2

x² + y² + 2by + c² = 0
x² + (y + b)² = b² - c²
The center is (0, -b) and radius is √(b² - c²)

The distance between two centers is same as sum of the radius of them:

d = √(a² + b²)

Sum of radiuses:

√(a² - c²) + √(b² - c²)

Since they are same we have:

√(a² + b²) = √(a² - c²) + √(b² - c²)

Square both sides:

a² + b² = a² - c² + b² - c² + 2√(a² - c²)(b² - c²)
2c² = 2√(a² - c²)(b² - c²)

Square both sides:

c⁴ = (a² - c²)(b² - c²)
c⁴ = a²b² - a²c² - b²c² + c⁴
a²c² + b²c² = a²b²

Divide both sides by a²b²c²:

1/a² + 1/b² = 1/c²

Proved

6 0

3 years ago

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