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2 years ago

Need answer ASAP I’m timed What is the distance between the points (7, 8) and (-8, 0) on a coordinate grid?

Mathematics

1 answer:

Whitepunk [10]2 years ago

3 0

Answer:

Step-by-step explanation:

(7, 8) (-8, 0)

First find how much the point moved in y to the second point.

8-0=8

So one side is 8.

Then find how much it moved in x. 7-(-8)=15

So the other side is 15

This will give you 2 sides of a triangle where one sides lenght is 8 the other is 15. Use the pythagoras theorem which is x squared plus y squared = c squared

8^2+15^2=c^2

64+225=c^2

289=c^2 Take the square root of both sides to solve for C

√289= √C^2

17 = c

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A garden in the shape of an equilateral triangle

Strike441 [17]

Answer:

area is $25\sqrt{3}$

Step-by-step explanation:

the hight is $5\sqrt{3}$

$5\sqrt{3}$ *10/2= $25\sqrt{3}$

8 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

Find the area of the shape shown below.

vlabodo [156]

Answer:

6 $cm^{2}$ (assuming the units is in centimetres)

Step-by-step explanation:

Hi, hope this helps!

Method 1:

Calculate this shape as a trapezium. This is a trapezium because it has one pair of parallel sides. The way to calculate the area of a trapezium is this:

- Half the sum of the parallel sides

- Multiply the perpendicular height between them

In this case, 4 and 2 are parallel, so we add them together (4 + 2 = 6) then we divide by two (6 ÷ 2 = 3). Then we multiply our answer by the perpendicular length between the two parallel sides (assuming the side on the left is at a right-angle, 3 x 2 = 6).

Method 2:

Calculate this shape as a rectangle + a triangle. If you split the shape so the pointy bit on the right becomes a triangle and the left becomes a square, you can calculate the area without knowing the formula for calculating a trapezium (see above ^).

- Area of the square / rectangle = length x width = 2 x 2 = 4

- Area of the triangle = 1/2 x length x width = 2 x 2 x 1/2 = 4 x 1/2 = 4 ÷ 2 = 2

Then we add the two areas together (4 + 2 = 6).

I hope this helped and I wasn't waffling on :)

Bluey

8 0

3 years ago

Read 2 more answers

4 x 4 5/6 simplify equation

mixas84 [53]

= 856

Step-by-step explanation:

im not sure :) ....

7 0

2 years ago

A jet flew from new york to london in 7 hours the average speed was 499.4 miles an hour how many miles was the trip

DENIUS [597]

499.4 x 7 = 3495.8 miles

4 0

3 years ago