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3 years ago

Solve the equation for y. 4x - 5y = 9

Mathematics

1 answer:

babymother [125]3 years ago

3 0

Answer:

y = (4x - 9) / 5 or y = 4/5x - 9/5

Step-by-step explanation:

Solving for y

4x - 5y = 9

-5y = -4x + 9

5y = 4x - 9

y = (4x - 9) / 5 or y = 4/5x - 9/5

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What is the measure for the following ∠? m∠BOA 20 40 60

kaheart [24]

Hello!

First, find angle BOA on the protractor. To do this, look for point B, point O, and point A.

Now, we see some markings that mark what degree point B and point A are at. Point B is marked at 60 degrees, and Point A 40 degrees.

Now, all that's left to do, is subtract the two measurements.

60 degrees - 40 degrees = 20 degrees

Therefore, your answer is 20.

Hope this helps!

8 0

3 years ago

a cone has a volume of 1230.88 units cubed and a diameter of 14 units what is the height of the cone use 3.14 for pi

ANEK [815]

Answer:

h= 24

Step-by-step explanation:

$v = \pi {r}^{2} \frac{h}{3}$

$r = \frac{d}{2}$

$r = \frac{14}{2}$

r= 7

1230.88=3.14×7^2×h/3

1230.88= 153.86h/3

h=1230.88/153.86

h= 24

7 0

2 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.

6 0

2 years ago

How does the slope of a line relate to real world situations

snow_lady [41]

I think construction jobs might use it

7 0

2 years ago

melamori03 [73]

Answer: 162.00 or 40.50 or 202.50

Step-by-step explanation:

$6.75 times 30 = 202.50

202.50 times 0.2 = 40.50

202.50 minus 40.50 = 162.00

7 0

3 years ago