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3 years ago

4. The sum of Mr. and Mrs. Bergen's ages is 100. The difference between their ages is 10. How old are Mr. and Mrs. Bergen?

Mathematics

1 answer:

alex41 [277]3 years ago

8 0

100-10=90

90÷2=45

Mr Bergen's age=45+10

=55 years old

Mrs Bergen's age=45 years old

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(1,2)(-2,6)
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Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

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Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

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3 years ago

Which rule describes a composition of transformations that maps pre-image PQRS to image P"Q"R"S"?

FinnZ [79.3K]

The correct option is Option D $\boxed{{r_{y-axis}}o{R_{0,270^\circ }}\left({x,y}\right)}$ .

Further explanation:

A translation is a transformation that transforms the figure with a fixed distance in the same direction.

A rotation is the transformation that rotates the figure with given angles.

Given:

It is given that the two transformations that maps pre-image PQRS to image ${\text{P''Q''R''S''}}$ .

Step by step explanation:

Step 1:

It can be seen from the given figure that that the pre image is in the first quadrant and the image is the third quadrant.

The coordinates in the second quadrant represents as $\left({-x,y}\right)$ and in the third quadrant represents as $\left({-x,-y}\right)$ if $x,y$ are positive.

Therefore, the rotation is in the counter clockwise direction of $270^\circ$ .

Step 2:

The rotation of $270^\circ$ in the counter clockwise direction represents the coordinates as,

$\left({x,y}\right)\to\left({y,-x}\right)$

It can be seen that the coordinate of ${\text{PQRS}}$ are as follows,

$\begin{aligned}P=\left({1,1}\right)\hfill\\Q=\left(1,5}\right)\hfill\\R=\left({3,5}\right)\hfill\\S=\left({3,1}\right)\hfill\\\end{aligned}$

Then after rotation of $270^\circ$ counterclockwise on ${\text{PQRS}}$ $\left( {x,y}\right)\to\left({y,-x}\right)$ as,

$\begin{gathered}P\left({1,1}\right)\to\left({1,-1}\right)\hfill\\Q\left({1,5}\right)\to\left({5,1}\right)\hfill\\R\left({3,5}\right)\to\left({5,-3}\right)\hfill\\S\left({3,1}\right)\to\left({1,-3}\right)\hfill\\\end{gathered}$

Step 3:

Now apply the rule of $y$ axis of reflection ${R_{y-axis}}\left({x,y}\right)\to\left({-x,y}\right)$ on the above transformation as,

$\begin{gathered}\left({1,-1}\right)\to\left({-1,1}\right)=P''\hfill\\\left({5,1}\right)\to\left({-5,-1}\right)=Q''\hfill\\\left({5,-3}\right)\to\left({-5,-3}\right)=R''\hfill\\\left({1,-3}\right)\to\left({-1,-3}\right)=S''\hfill\\\end{gathered}$

Therefore, the given transformation is the rotation of $270^\circ$ counterclockwise followed by $y$ axis of reflection.

Therefore, this is the composition of transformation.

The composition of the given transformation can be written as,

${r_{y-axis}}o{R_{0,270^\circ}}\left({x,y}\right)$

Therefore, option D ${r_{y-axis}}o{R_{0,270^\circ}}\left({x,y}\right)$ is correct.

Learn more:

Learn more about what is the final transformation in the composition of transformations that maps pre-image abcd to image a"b'c"d"? a translation down and to the right a translation up and to the right a 270° rotation about point b' a 180° rotation about point b' brainly.com/question/2480946

Learn more about the transformation of function brainly.com/question/7297858

Learn more about midpoint of the segment brainly.com/question/3269852

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Transformations

Keywords: transformations, dilation, translation, rotation, counterclockwise, angle, clockwise, coordinates, mapping, rigid transformation, right side, left side, quadrant, composition.

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4. The sum of Mr. and Mrs. Bergen's ages is 100. The difference between their ages is 10. How old are Mr. and Mrs. Bergen?​

4. The sum of Mr. and Mrs. Bergen's ages is 100. The difference between their ages is 10. How old are Mr. and Mrs. Bergen?