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matrenka [14]
3 years ago
10

Can anyone help me out

Mathematics
1 answer:
choli [55]3 years ago
4 0
The height(zipper) would be 3 feet or 36 inches
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10. Make q the subject of the formula 5(q + p) = 4 + 8p
vichka [17]

The formula subject to q is \frac{4+3p}{5}

Explanation:

The given formula is 5(q+p)=4+8 p

We need to determine the formula subject to q.

<u>The formula subject to q:</u>

The formula subject to q can be determined by solving the formula for q.

Let us solve the formula.

Thus, we have;

5q+5p=4+8p

Subtracting both sides by 5p, we have;

5q=4+3p

Dividing both sides by 5, we get;

q=\frac{4+3p}{5}

Thus, the formula subject to q is \frac{4+3p}{5}

5 0
2 years ago
A train travels 220 miles in 4 hours. What is the train's speed in feet per minute?
Katena32 [7]
220 miles in 4 h = 55 mph multiply this by 5280 to get feet per hour then divide this by 60 to get ft/min


7 0
3 years ago
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WILL MARK BRAINLEIST!!!!! HELP ASAP!
Alex Ar [27]

Answer:

1 7/5

Step-by-step explanation:

we should first simplify the fraction part of each option

7/5 = 1 2.5 add this to 1 and you get 2 2/5

this happens to be the answer but in the future go through each option

5 0
3 years ago
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7⋅5+(4^2−2^3)÷4 answers: 49 41 34 9
Bess [88]

Answer: 37

Step-by-step explanation:

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2 years ago
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Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp
olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = \overline{x} = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data​ speeds i.e. x - \overline{x}

x - \overline{x} = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

\frac{59.98}{20.03}=2.99

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

z=\frac{x-\overline{x}}{s}

Using the given values, we get:

z=\frac{75.6-15.62}{20.03}=2.99

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly​ high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the  carrier's highest data speed​ is significantly higher.

3 0
3 years ago
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