All categories

4 years ago

Dividing my dog’s age by 4 and adding 6 is the same as dividing my dog’s age by 8 and adding 8.

Mathematics

1 answer:

sergiy2304 [10]4 years ago

4 0

Answer:

Step-by-step explanation:

a = age

$\frac{a}{4} + 6 = \frac{a}{8} + 8$
you will multiply by 8 now
$2a + 48 = a + 64$
now move a's with a's and the numbers that don't have a variable with each other
$2a - a = 64 - 48$
now subtract
$a = 16$
I hope it helps you!

You might be interested in

What is the measure of the missing angle?

nekit [7.7K]

Answer:

Step-by-step explanation:

A triangle always measures to be 180 degrees. 50+50=100 so 80 degrees is left

5 0

3 years ago

Read 2 more answers

A shopkeeper wants to make identical fruit baskets out of 32 apples, 40 peaches, and 96 bananas. What is the largest number of b

Finger [1]

4 baskets as the largest number divisible by 32 is 8 and the other 2 are also divisible by 8

7 0

4 years ago

HELLPP PLZZ Which value from the set {3, 9, 24, 36} makes m(2^2 + 2^3 ) = 3 • 12 true?

Stells [14]

Answer:

12*3 is 36 and (2^2 + 2^3 ) is 12 so you have to get 12 to 36

Step-by-step explanation:

which is 12 *3 so 3

7 0

3 years ago

59 = x-10 59+ __ = X-10 + = x 59 = x - 10 59+ = x - 10+ = x

8_murik_8 [283]

Answer:

$59 = x - 10 \\ 59 + 10 = x \\ x = 69$

4 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

3 years ago