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Alexus [3.1K]
3 years ago
7

14) Find the slope of the line containing the points (5, 3) and (-7, 2).

Mathematics
1 answer:
Zina [86]3 years ago
7 0

Answer:

1/12

Step-by-step explanation:

Step one:

given data

the line containing the points (5, 3) and (-7, 2).

x1= 5

y1= 3

x2= -7

y2=2

Step two:

The slope is given as

Slope= y2-y1/x2-x1

substitute

Slope= 2-3/-7-5

Slope= -1/-12

Slope = 1/12

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Each week, Rosario drives to an ice-skating rink that is 60 miles away. The round-trip takes 2.75 hours. If he averages 55 miles
sertanlavr [38]

Answer:

60(55)•60x=2.75

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
A bag contains 10 marbles:
Len [333]
Total number of marble = 10
3green
2red
5 blue
Probability that the first marble is red = 2/10

Probability that the second is blue = 5/9(a reduction in the total number of marbles, because after the first marble was picked it wasn't replaced)

Probability of 1st and 2nd being red and blue respectively = 2/10 × 5/9
=1/9....

Hope this helped...?

6 0
3 years ago
Researchers fed mice a specific amount of Dieldrin, a poisonous pesticide, and studied their nervous systems to find out why Die
Elodia [21]

Answer:

Step-by-step explanation:

Part A

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

6 0
3 years ago
Whilst shopping, the probability that Caroline buys fruit is 0.4. The probability that Caroline independently buys a CD is 0.2.
AleksandrR [38]

Answer:

probability that Caroline buy both CD and  fruit = 0.52

Step-by-step explanation:

We have given that the probability of Caroline buys a fruit P = 0.4

So probability of Caroline does not buy the fruit = 1 - 0.4 = 0.6

Probability Caroline buys a CD P = 0.2

So probability of Caroline does not buy the CD = 1 - 0.2 = 0.8

So probability that Caroline does not buy either buy CD or fruit = 0.8×0.6=0.48

So probability that Caroline buy both CD and  fruit =1-0.48 = 0.52

7 0
3 years ago
Please help me with this please and thank you
melamori03 [73]

Answer:

B

Step-by-step explanation:

let X represent the smaller paintings and Y represent the larger paintings

since she bought 8 paintings

X+Y=8

She bought all the eight paintings for $230 and X is worth 25 while Y is worth 40

25×X+40×Y=230

25X+40Y=230

7 0
2 years ago
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