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3 years ago

Can someone plz help I put the picture

Mathematics

1 answer:

Svetlanka [38]3 years ago

3 0

Answer:

personally i think x=6

Step-by-step explanation:

but heres how i got it cross multiiply 7.5/4 and x/3.2

witch makes 24=4x then divide by 4 and you get x=6

so.... yeah i think x=6

if brainiest is earned its greatly appreciated

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A fair die is tossed 5 times. Let <img src="https://tex.z-dn.net/?f=%5Cdfrac%7Bm%7D%7Bn%7D" id="TexFormula1" title="\dfrac{m}{n}

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Answer:

m = 1

Step-by-step explanation:

We can suppose that the number we are looking for is for example 5.

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For the first toss the probability we have 5 is 1/6 (we have 6 numbers on the dice and number 5 is just one of the possible 6 outcomes).

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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.

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3 years ago