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vichka [17]
3 years ago
5

Omealio received a grade of 80 on Exam 1. If his grade in Exam 2 was 15% higher, what was his grade in Exam 2? Please help

Mathematics
2 answers:
abruzzese [7]3 years ago
8 0

Answer:

92

Step-by-step explanation:

15 percent of 80 is 12 percent

80 + 12 = 92

ExtremeBDS [4]3 years ago
7 0

Answer:

well if the max grade he could get was 100 then you need to find 15% of 100 which is 15, after that add 15 to 80 and you get 95.

his grade for exam 2 was 95

Step-by-step explanation:

15% of 100 = 15

80 + 15 = 95

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Let f (x) = 2x - 1, g(x) = 3x, and h(x) = x2 + 1. Compute the following: h (h(x)) and. g (h (x))
ki77a [65]
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7 0
2 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
I need help with number 10 please I’m so confused on how to do it!
Solnce55 [7]

Given:

A figure.

m \angle C D A=9 x+3 and m \angle E D A=6 x+5

To find:

What kind of figure and the value of x

Solution:

All four sides are congruent.

Diagonals bisect each other.

There the given figure is rhombus.

Diagonals bisect the angles.

⇒ m \angle C D A=m \angle E D A

9 x+3=6x+5

Subtract 3 on both sides.

9 x+3-3=6x+5-3

9 x=6x+2

Subtract 6x from both sides.

9 x-6x=6x+2-6x

3x=2

Divide by 3 on both sides.

$\frac{3x}{3} =\frac{2}{3} $

$x=\frac{2}{3} $

The value of x is \frac{2}{3}.

5 0
3 years ago
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