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djverab [1.8K]
3 years ago
13

What percent is equivalent to 0.781

Mathematics
1 answer:
JulijaS [17]3 years ago
8 0
78.1% or 78% rounded
You might be interested in
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
Is 36 the median of 35,41,18,75,36,21,62,29,154,70
irga5000 [103]

Answer:

no, it is 38.5

Step-by-step explanation:

you have to add up the two numbers that are in the middle then divide it by two since there isn't just one number in the middle.

6 0
3 years ago
Suppose f(x)=x^2 +1. Find the graph of 4 f(x). <br> Graph 1<br> Graph 2
umka21 [38]

Answer:

Graph 1.

Step-by-step explanation:

The graph of f(x) will be compressed by a factor 4 and the vertex will be 4 units up from the origin.

6 0
3 years ago
Which angles are supplementary to each other?
Nat2105 [25]

Answer:

dea/ced

Step-by-step explanation:

8 0
4 years ago
Read 2 more answers
Triangle JKM with side j across from angle J, side k across from angle K, and side m across from angle M
Virty [35]

j=43.8 ft

Step-by-step explanation:

Given that ∠K =110° and corresponds to side k=70 ft, then applying the laws of sines then

k/sine∠K =j/sine ∠J

70/sine 110° = j/ sine 36°

j= 70 sin 36° / sin 110°

j=43.785

j=43.8 ft

Learn More

Sine rule :brainly.com/question/9915319

Keywords: triangle, sides,angle,law of sines,nearest tenth

#LearnwithBrainly

7 0
3 years ago
Read 2 more answers
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