What is the scale factor of point E (original point) to point F (new point) with P as the center of dilation.

In autumn a tree starts to shed its leaves. 2/5 of the leaves fall off in the first week. 1/2 of those remaining fall of in the

Oduvanchick [21]

Answer:

850 leaves were on the tree initially.

Same as;

2/5 + 1/5 + 3/10 + (remaining leaves 1/10) = 1

4/10 + 2/10 + 3/10 + 1/10

10 x 85 = 850

Step-by-step explanation:

2/5 wk 1 fall off

1/5 wk 2 fell off

3/5 x 1/2 = 3/10 wk 2 stayed on tree

1/3 x 3/10 = 1/10 wk 3 stayed on

2/5 + 1/5 + 3/10 = 9/10 = 0.9 fall off in total

10/10 - 9/10 = 1/10 left on tree

1/10 = 85

10 x 85 = 850 leaves.

8 0

2 years ago

For any two integers a and b, (a + b)² = a² + b²

kozerog [31]

Answer:

wrong

Step-by-step explanation:

(a+b)^2=(a+b)(a+b)

=a(a+b)b(a+b)

=a^2+ab+ab+b^2

=a^2+2ab+b^2

i hope it helps you

3 0

3 years ago

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

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 ] .

4 0

2 years ago

Answer this question for brainliest and 20 points

Naddik [55]

Answer:

Least to greatest: 2, 3, 1

Step-by-step explanation:

Square root of 8 over 2 is about 1.41 which makes it the least value

2 is obviously equal to 2 which puts it in the middle

Square root of 7 is about 2.64 which makes it the greatest value

3 0

2 years ago

Alexis has a pocket full of coins that consists of dimes and nickels. She has a total of 40 coins. The total value of her coins

MrMuchimi

Answer

3 quarters 7 dimes

Step-by-step explanation:

4 0

3 years ago