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3 years ago

Which box plot represents the dot plot data

Mathematics

1 answer:

ycow [4]3 years ago

7 0

Hey!
The median is 28, so it isn’t A or B. I can’t see C and D very well but if C’s median is 30 ( and I think that’s what it says) then it has to be D.
Hope this helps!

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Use the equation p=10j to find the value of p when j=6.

lyudmila [28]

Answer:

p=60

Step-by-step explanation:

10(6)=60

p=60

Please mark as Brainliest! :)

Have a nice day.

7 0

3 years ago

Read 2 more answers

Paula's Flowers sells 39 arrangements per week. The flower shop down the street sells

Eddi Din [679]

Answer:

50:39

Step-by-step explanation:

The first number in the context always goes first!

4 0

3 years ago

Number 4 please. its confusing

Deffense [45]

Answer:

63 monocotyledons

Step-by-step explanation:

some abbreviations ill be using:

monocotyledons=m

dicotyledons=d

3m / 4d

x / 84d

you need to figure out what x is

you can divide 84/4 and get 21

now that you have 21, you can multiple that number by 3

21x3=63

therefore, there are 63 monocotyledons

7 0

3 years ago

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starting at home, Omar traveled uphill to the gift store for 30 minutes at just 10mph. He then travelled back home along the sam

SashulF [63]

Answer:

average speed for the entire trip is 15 mph

Step-by-step explanation:

Let x be the uphill speed and y be the downhill speed.

We have been given that

Speed of Omar in uphill to reach gift store = x =10 mph

Speed of Omar in downhill to reach his home = y = 30 mph

We know the formula for average speed for the entire trip

$\text{Average speed }=\frac{2xy}{x+y}\\\\=\frac{2\times 10\times 30}{10+30}\\\\=\frac{600}{40}\\\\=15$

Therefore, average speed is 15 mph

6 0

3 years ago

Read 2 more answers

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

3 years ago