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Travka [436]
1 year ago
5

WILL GIVE BRAINLEST!!!

Mathematics
1 answer:
melomori [17]1 year ago
6 0
Sounds like answer d to me
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Are there any linear quantities in the chart shown? And if so, what is the slope of said pairs?
VashaNatasha [74]

The chart does not represent a linear quantity

<h3>How to determine if there are linear quantities in the chart shown?</h3>

From the chart, we have the following coordinates

(x, y) = (0.10, 0.14), (0.50, 0.32), (1.00, 0.46), (1.70, 0.59) and (2.00, 0.63)

To determine if the chart is a linear quantity, we simply calculate the slope using the coordinates.

The slope is calculated as:

m = (y2 - y1)/(x2 - x1)

So, we have:

m = (0.32 - 0.14)/(0.50 - 0.10) = 0.45

m = (0.46 - 0.32)/(1.00 - 0.50) = 0.28

In the above computations, the calculated slopes are not equal

Hence, the chart does not represent a linear quantity

Read more about linear equations at:

brainly.com/question/1884491

#SPJ1

7 0
2 years ago
Over the summer, for every 14 okra seeds Dana planted, 9 grew into plants. If he planted 210 okra seeds, how many grew into plan
Katena32 [7]
210s(9p/14s)=135p

So 135 seeds grew into plants.
8 0
3 years ago
Read 2 more answers
PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST
ohaa [14]

Answer:

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.

Step-by-step explanation:

7 0
3 years ago
the first Indianapolis 500 auto race took place in 1911. the winning car covered the 500 miles in 6.7 hours. what was the winnin
Vesna [10]

Average speed = (distance covered) / (time to cover the distance)

                         =        (500 miles)      /    (6.7 hours)

                         =            (500 / 6.7)  (mile/hour)

                         =                    74.63 miles per hour.

I imagine there were quite a number of pit stops included in that.

This is faster than I expected when I first read your question.
That's really gettin' with it for cars in 1911 !

4 0
3 years ago
Quick anyone know this algebra 2 problem?
anastassius [24]

I believe it's the 3rd one

3 0
3 years ago
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