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pychu [463]
3 years ago
8

What adds up to 14 and multiplies to negative 576?

Mathematics
2 answers:
kkurt [141]3 years ago
7 0
24 x 24 = 576 
<span>24 + 24 = 48 </span>

<span>576 is equal to 24 squared. Hope i helped ;)</span>
DedPeter [7]3 years ago
3 0
32 and -18 add to 14 and multiply to negative 576
You might be interested in
Question 5 Unsaved
Otrada [13]

We can use the compound interest formula

F=P(1+i)^n

where

F=Future value of investment to be found

P=present value of investment ($1000)

i=interest per period (1/4 year)=0.04/4=0.01

n=number of periods (3 years * 4 quarters = 12)


Substitute or "Plug in" values, so to speak,

F=1000*(1+0.01)^12

use a calculator to do the sum

=1126.83 (to the nearest cent, and use the proper rounding rules)



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
3 years ago
Give the digits in the ones place and the tenths place.<br> 86.59
mr_godi [17]
Digit in ones place- 6
Digit in tenths place- 5
4 0
4 years ago
Veronica has been saving dimes and quarters and has a total of 86 coins worth $9.00. How many
taurus [48]
The answer to your question is 4.11 if I’m correct
8 0
3 years ago
Please answer this question as soon as possible
Taya2010 [7]

Answer: The graph is shifted vertically, but not horizontally

Step-by-step explanation:

Since the added constant is outside the x term, the constant will only shift the graph vertically.

6 0
3 years ago
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