If point F is translated 2 units to the left and 2 units down, what are the coordinates of F'?

A^2-b^2 a^2b-ab^2 -------------- ÷ ---------------a^2b+ab^2 a^2b- ab^2

Damm [24]

Answer:

$\huge{= \frac{ {a}^{2} - {b}^{2} }{ {a}^{2b} + {ab}^{2} }}$

Step-by-step explanation:

$\huge{ \frac{ {a}^{2} - {b}^{2} }{ {a}^{2b} + {ab}^{2} } \div \frac{ {a}^{2b} - {ab}^{2} }{ {a}^{2b} - {ab}^{2} }}$

$\huge{\frac{ {x}^{2} - {b}^{2} }{ {a}^{2b} + {ab}^{2} } \div 1}$

$\huge{\boxed{\green{= \frac{ {a}^{2} - {b}^{2} }{ {a}^{2b} + {ab}^{2} }}}}$

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.

6 0

3 years ago

23. If 25% of a number is 75. find the number. A. 700 B. 600 C. 500 D, 400 E. 300

nadya68 [22]

Answer:

23.E

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

Se vendieron un total de 430 boletos para la obra de teatro escolar. Los boletos eran o de adulto o de estudiante. Se vendieron

BartSMP [9]

Resolviendo un sistema de ecuaciones, veremos que se vendieron 250 boletos de adulto.

<h3>¿Cuántos boletos de adulto fueron vendidos?</h3>

Definamos las variables:

x = boletos de adulto vendidos.
y = boletos de estudiantes vendidos.

Se vendieron un total de 430 boletos, entonces:

x + y = 430

Y se vendieron 70 boletos de estudiante menos que boletos de adulto:

y = x - 70

Tenemos el sistema de ecuaciones:

x + y = 430

y = x - 70

Para resolverlo, podemos reemplazar la segunda ecuación en la primera:

x + y = 430

x + (x - 70) = 430

2x = 430 + 70 = 500

x = 500/2 = 250

Se vendieron 250 boletos de adulto.

Sí quieres aprender más sobre sistemas de ecuaciones:

brainly.com/question/13729904

#SPJ1

5 0

2 years ago

The is a function of the

Delvig [45]

Answer:

Is this the right question?

Step-by-step explanation:

3 0

3 years ago