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3 years ago

help with this please

Mathematics

1 answer:

Ierofanga [76]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Under a translation < 5, - 9 >

5 is added to the original x- coordinate and 9 is subtracted from the original y- coordinate, that is

A(1, 4 ) → A'(1 + 5, 4 - 9 ) → A'(6, - 5 )

B(2, - 2 ) → B'(2 + 5, - 2 - 9 ) → B'(7, - 11 )

C(- 3, 2 ) → C'(- 3 + 5, 2 - 9 ) → C'(2, - 7 )

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What is the value of x in the equation 7x + 2y = 48 , when y = 3?

9966 [12]

7x + 2y = 48
7x + 2(3) = 48
7x + 6 = 48
7x = 42
x = 6

answer when y = 3, x = 6

7 0

3 years ago

Find the missing number.

DENIUS [597]

Answer:

20%

Step-by-step explanation:

In the graph, there is a cycle of increasing and decreasing. First, the original circle increases by 25% into the new, larger circle. Then, the circle does back to its original size. So, to solve we can assign a value to the smaller circle, increase it by 25%, and then find what percentage brings this back to the original value.

It is easiest to assign the value of 100 because it is easy to work with. So, let's say the smaller circle is 100; this increased by 25% is 125. The next step is to find what percentage of 125 brings us back to 100. There are multiple ways to do this such as guess and check or proportions/fractions. Either way, the correct percentage is 20%. If you decrease 125 by 20% you get the correct value of 100. This means that the missing number must be 20%.

6 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.

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3 years ago

First discovered more than 30 years ago, Lina’s sunbird, a four-and-a-half-inch animal found in the Philippines and that resembl

Arlecino [84]

Answer: the correct one is C) found in the Philippines and resembling

Step-by-step explanation:

To maintain parallelism, you must list items with equal grammatical structure.

"Found" is just an adjective that modifies the animal, and so is "resembling." Sure, they don't sound the same, but they're both adjectives.

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4 years ago

Rita is making chili.The recipe calls for 2 3/4 cups of tomatoes. How many cups of tomatoes, written as a fraction greater than

Andrei [34K]

The way the this is worded means to turn it into a improper fraction which means the answer is 11/4

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3 years ago

￼help with this please

help with this please