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3 years ago

Y - 4 × x - 6 is?

Mathematics

2 answers:

Vlad [161]3 years ago

6 0

Answer:

y-4x-6

Step-by-step explanation:

I'm not 100% sure but I think this is correct!

Strike441 [17]3 years ago

6 0

Answer:

y - 4x - 6

Step-by-step explanation:

y - 4 • x - 6 = y - 4x - 6

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Write the equation of the graph in vertex form enter power using ^ and don’t forget y or f(x)

enyata [817]

Me!!!! 245554! Then do 45316

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3 years ago

The moon is a sphere with radius of 959 km. Determine an equation for the ellipse if the distance of the satellite from the surf

Sergeeva-Olga [200]

Answer:

$\frac{x^2}{1316^2}+\frac{y^2}{1669^2}=1$

Step-by-step explanation:

An ellipse is the locus of a point such that its distances from two fixed points, called foci, have a sum that is equal to a positive constant.

The equation of an ellipse with a center at the origin and the x axis as the minor axis is given by:

$\frac{x^2}{b^2}+\frac{y^2}{a^2} =1 \\\\where\ a>b$

Since the distance of the satellite from the surface of the moon varies from 357 km to 710 km, hence:

b = 357 km + 959 km = 1316 km

a = 710 km + 959 km = 1669 km

Therefore the equation of the ellipse is:

$\frac{x^2}{1316^2}+\frac{y^2}{1669^2}=1$

5 0

3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

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3 years ago

What number does these numbers share -21,22,65?

finlep [7]

Answer:

uh cuz

Step-by-step explanation:

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3 years ago

a large serving of soup from Dayton's restaurant is 4/5 liters. The restaurant has 4 liters of soup leftover at the end of the n

DochEvi [55]

The restaurant has 5 servings left.
All you do is multiply 4 by 1.2

5 0

3 years ago

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Y - 4 × x - 6 is? ​

Y - 4 × x - 6 is?