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geniusboy [140]
3 years ago
9

Please help me out? thank you.

Mathematics
1 answer:
anzhelika [568]3 years ago
4 0
It’s the fourth answer
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Thrice a number decreased by 5 exceeds twice the number by a unit find the number
viktelen [127]
Working attached below. Hope this helps! Any questions let me know :)

6 0
3 years ago
The price of an item has been reduced h 45% the original price was $49
borishaifa [10]

Answer:

$26.95, you save  $22.05

Step-by-step explanation:

$49 x 0.45 = $22.05 savings, 49 - 22.05 = 26.95

3 0
3 years ago
Martha and sarah leave block p and q towards q and p respectively simultaneously and travel in the same route. After meeting eac
olga_2 [115]

The average speed of Martha and Sarah is 32 km/h.

We need to know about the speed to solve this problem. Speed can be determined as the distance traveled divided by time. It can be written as

v = s / t

where v is speed, s is distance and t is time.

From the question above, we know that:

t sarah = 3 hours

t martha = 5 hours

v sarah = 40 km/h

By using the speed equation, we get the distance

vsarah = s / tsarah

40 = s/3

s = 120 km

Find Martha's speed

vmartha = s / tmartha

vmartha = 120 / 5

vmartha = 24 km/h

Find average speed

v = (vsarah + vmartha)/2

v = (40 + 24) / 2

v = 32 km/h

Hence, the average speed of Martha and Sarah is 32 km/h.

Find more on speed at: brainly.com/question/6504879

#SPJ4

5 0
2 years ago
NEED HELP ASAP
Lostsunrise [7]

Answer:

0.09

step by step explanation

7 0
3 years ago
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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