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ella [17]
3 years ago
7

If a car is traveling at 65 miles per hour, then how far does it travel in 2 hours?

Mathematics
2 answers:
frez [133]3 years ago
5 0
65 miles per hour means that for each hour, it goes 65 miles

65 miles*2 hours= 130 miles

Final answer: 130 miles
Sidana [21]3 years ago
4 0
If it travels 65 miles for on hour, you just need to double that amount to get 2 hours, so 65+65 is 130. 130 miles in 2 hours.
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Part A. Write the equation 5 = 10x – 5y in slope-intercept form.
bagirrra123 [75]

Answer:

Slope intecept form: y=2x-1

Y-intercept: (0,-1)

and the slope is 2 if you nee it.

4 0
2 years ago
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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
Using transformation (x-2 y+3) find the new coordinates of (7 -5) brainly
trasher [3.6K]

Answer:

(5, -2)

Step-by-step explanation:

In the coordinates (7, -5), 7 is the x-coordinate and -5 is the y-coordinate.

The transformation, (x-2 y+3), states that the x-coordinate, 7, must be subtracted by 2.

When subtracted by two, (7 - 2), the difference is 5.

The transformation, (x-2 y+3), states that the y-coordinate must be increased by 3.

When added by 3, (-5 + 3), the sum is -2.

Therefore, the new coordinates are (5, -2).

3 0
2 years ago
4/5 dived by 3/4<br><br> Please help
Leto [7]

Answer:

1 1\15

Step-by-step explanation:

5 0
3 years ago
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Urgent math help needed will mark brainliest
iren [92.7K]

Remark

Many calculators have a key that will do this in one step. Mine is one of them. So let's start with the answer and then we'll do it the long way.

9C3 = 84. You put it into your calculator as 9   2nF   nCr 3 =

Solution

\dfrac{9!}{(n - 3)!3!} = \dfrac{9*8*7*6!}{6!*3!}=\dfrac{9*8*7}{6}

3*4*7 = 84 which is what you set out to show.


3 0
3 years ago
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