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3 years ago

The point (1, 4) is translated so it ends on the coordinates (-2, 2). Which of the following describes the translation?

Mathematics

1 answer:

Rom4ik [11]3 years ago

7 0

Answer:

The first answer choice is the correct one.

Step-by-step explanation:

Going from (1, 4) and ending up at (-2, 2), we see that the original point is now 3 units to the left of its starting position, and 2 units down.

Clarification: from x = 1 to x = -2 is 3 units (to the left), and

from y = 4 to y = 2 is 2 units (down)

The first answer choice is the correct one.

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Proving trapezoid theorems Geometry

Artist 52 [7]

Answer:

ΔABD ≅ ΔACD by SAS, therefore;

$\overline {BD} \cong \overline {CA}$ by CPCTC

Step-by-step explanation:

The two column proof is presented as follows;

Statement ${}$ Reason

ABCD is a trapezoid ${}$ Given

$\overline {BA} \cong \overline {CD}$ ${}$ Given

$\overline {BC} \parallel \overline {AD}$ ${}$ Definition of a trapezoid

ABCD is an isosceles trapezoid ${}$ Left and right leg are equal

∠BAD ≅ ∠CDA ${}$ Base angle of an isosceles trapezoid are congruent

$\overline {AD} \cong \overline {AD}$ ${}$ Reflexive property

ΔABD ≅ ΔACD ${}$ By SAS rule of congruency

$\overline {BD} \cong \overline {CA}$ ${}$ CPCTC

CPCTC; Congruent Parts of Congruent Triangles are Congruent

SAS; Side Angle Side rule of congruency

8 0

3 years ago

Read 2 more answers

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

3 years ago

In a certain elementry school, 52% of the students are girls. a sample of 65 students is drawn. what is the probability that mor

Rainbow [258]

Answer:

12432435

Step-by-step explanation:

gvngxsfgjghdjmhg

7 0

3 years ago

Read 2 more answers

Please help! I’ve been trying to find the answer to this problem and couldn’t & I can’t find the answers anywhere!

Kryger [21]

The answer is if you tilt it a little bit it looks like a kite and if you look on the web a rhombus kinda looks like a kite and plz mark me brainiest because you know you looked for answers on the web

6 0

3 years ago

Adult tickets for a concert cost $16.00, and student tickets cost $12.00. If twice as many student tickets as adult tickets were

Len [333]

Answer:

670 adult tickets
1340 student tickets

Step-by-step explanation:

One group of 2 student and 1 adult tickets will go for 2×$12 +16 = $40. The number of such groups sold was ...

$26,800/$40 = 670

There were 670 adult tickets sold.

There were 670×2 = 1340 student tickets sold.

6 0

3 years ago

Read 2 more answers