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fenix001 [56]
1 year ago
6

The train station clock runs too fast and gains 7 minutes every 4 days. How many minutes and seconds will it have gained at the

of 10 days
Mathematics
1 answer:
dolphi86 [110]1 year ago
7 0
4 days 10 days Cross Multiplication
---------- = ------------
7 mins x

4x = 7*10
4x = 70
x = 70/4
x = 17.5

17.5 minutes = x
You might be interested in
The New Haven High School cheerleaders got a t-shirt gun. The cheerleaders stand
nevsk [136]

Answer: the gun was held 4ft high i.e H = 4ft

Step-by-step explanation:

To determine how high the gun was held when it was fired H;

Before firing the t-shirt up, the gun and the t-shirt are both at the same height.

Therefore,

H = h(0) i.e h(t) at t = 0

h(t) = -16t^2 + 64t + 4

H = h(0) = -16(0)^2 + 64(0) + 4

H = 4ft

7 0
2 years ago
the distribution of scores on a recent test closely followed a normal distribution wotb a mean of 22 and a standard deviation of
soldi70 [24.7K]

Answer:

1) 22.66%

2) 20

Step-by-step explanation:

The scores of a test are normally distributed.

Mean of the test scores = u = 22

Standard Deviation = \sigma = 4

Part 1) Proportion of students who scored atleast 25 points

Since, the test scores are normally distributed we can use z scores to find this proportion.

We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:

P(X ≥ 25)

We can convert this value to z score and use z table to find the required proportion.

The formula to calculate the z score is:

z=\frac{x-u}{\sigma}

Using the values, we get:

z=\frac{25-22}{4}=0.75

So,

P(X ≥ 25) is equivalent to P(z ≥ 0.75)

Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266

Since,

P(X ≥ 25) = P(z ≥ 0.75), we can conclude:

The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%

Part 2) 31st percentile of the test scores

31st percentile means 31%(0.31) of the students have scores less than this value.

This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.

Using the z table, the z score for 31st percentile comes out to be:

z = -0.496

Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.

Using the values, we get:

-0.496=\frac{x-22}{4}\\\\ x=4(-0.496) + 22\\\\ x=20.02\\\\ x \approx 20

Thus, a test score of 20 represent the 31st percentile of the distribution.

3 0
2 years ago
Louden County Wildlife Conservancy counts butterflies each year. Data over the last three years regarding four types of butterfl
MrRa [10]

Incomple question. However, here's the remaining part of the question:

14

2009

Meadow Fritillary= 5

Variegated Fritillary= 7

Zebra Swallowtail= 33

Eastern-Tailed Blue= 242

Louden County Butterfly Count

2010

Meadow Fritillary 34

Variegated Fritillary 95

Zebra Swallowtail 21

Eastern-Tailed Blue 168

2011

Meadow Fritillary

Variegated Fritillary

Zebra Swallowtail

Eastern-Tailed Blue

10

170

<u>Options</u>:

A) All butterfly populations are steadily decreasing.

B)All butterfly populations were larger than usual in 2010.

C)The Eastern-Tailed Blue butterfly is more common than the others.

D)The Meadow Fritillary is equally common as the Variegated Fritillary

Answer:

<u>C</u>

Step-by-step explanation:

Looking through the above count data by Louden County Wildlife Conservancy from 2009 to 2011 we notice the Eastern- Trailed Blue butterfly has a higher count, which implies that the Eastern-Tailed Blue butterfly is more common than the other butterflies.

Therefore, we could infer from the samples, that the Eastern-Tailed Blue butterfly is more common than others from the records of the past 3 three years.

5 0
1 year ago
A. If a spacecraft was parked on Venus and needed to make a flight to Jupiter, how far would it need to travel? Show your work a
bija089 [108]
A. (Assume both planets are aligned with the sun and are on the same side of the sun.) Show your work and provide your answer in scientific notation. Jupiter = 778,500,000 km from sun. Venus = 108,200,000 km from sun.

778,500,000 – 108,200,000 = 670,300,000
It would need to travel 670,300,000 km/ 6.703 x 10 with a 8 squared km (as shown below)
10 {}^{8}


B.Mercury, Venus, and Earth are the three planets closest to the sun. Would their combined distance from the sun be greater or less than the distance from the sun to Neptune? Show your work and justify your answer. Mercury = 57,909,000 km Venus = 108,200,000 km Earth = 149,600,000 km 57,909,000 + 108,200,000 + 149,600,000 = 315,709,000 Less because 315,709,000 km is less than 4,498,000,000 km

C.If Earth was 10 times farther away from the sun than it is now, which planet would it be closest to? (Assume all the planets are aligned with the sun and are on the same side of the sun.) Compare Earth's new distance to that planet. How far apart would they be in standard notation? How far apart in scientific notation? Show your work. Earth = 149,600,000 km 149,600,000 x 10 = 1,496,000,000 Earth’s new distance = 1,496,000,000 Earth would be closest to Saturn. 1,496,000,000 – 1,429,000,000 = 67,000,000
They would be 67,000,000 km apart
Scientific notation: 6.7 x 10 with a 7 squared km (sorry I can't write this to show this figure but see below as you should write it exactly like how I typed it below followed by km)
10 {}^{7}


D. The space shuttle travels at about 28,000 km per hour. Using that information, estimate how many hours it will take the shuttle to reach Saturn from Earth. (Assume both planets are aligned with the sun and are on the same side of the sun.) Show your work. Convert your answer into scientific notation if necessary. Earth = 149,600,000 km
Saturn = 1,429,000,000 km
1,429,000,000 - 149,600,000 = 1,279,400,000
1,279,400,000 km
1,279,400,000 ÷ 28,000 = 45,692.8
Estimated hours it will take the shuttle to reach Saturn from Earth = 45,693
Scientific notation: 4.5693 x 10 with a 4 squared
(same with this one, it should be written as I have it shown below instead of how I first tried to explain it.)
10 {}^{4}






4 0
2 years ago
Biologists conducted a study to investigate the flying velocity of mosquitoes both before and after feeding. The following scatt
Lesechka [4]

Answer:

It represents a mosquito that flew very fast after feeding relative to all other mosquitoes.

Step-by-step explanation:

The point is unusual because the velocity of the mosquito is substantially greater than all other velocities and is probably an outlier.

4 0
1 year ago
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