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kow [346]
3 years ago
15

Does anyone know why x is normally used for an unknown number?

Mathematics
1 answer:
jeka943 years ago
7 0

Then you are able to substitute in X and it can have a infinite number of values catered to the question.

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Chris has finished 30% of an art project that has taken him a total of 9 hours so far. If he continues to work at the same rate,
Sveta_85 [38]
There was already a question like this on here so i’m going off the answer already given: 30 hours, i hope it’s right
3 0
2 years ago
Read 2 more answers
joey ate 1/2 of the pie. molly ate 1/2 of what was left. lola ate 1/2 of what was lefet after molly finished. what fractional pa
anzhelika [568]
Molly ate 1/4 and then the last one is 1/8
8 0
3 years ago
Alex and Freddy collect stamps. The number of stamps in Alex’s collection is in ratio 3:5 to the number of stamps in Freddy’s co
Tems11 [23]
Suppose Alex has 3x stamps, Freddy has 5x stamps. After Freddy gives Alex 10 stamps, Alex has 3x+10 while Freddy has 5x-10
the new ratio is 7 to 9, so we have the equation: (3x+10)/(5x-10)=7/9
9(3x+10)=7(5x-10)
27x+90=35x-70
160=8x
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So Alex has 3*20=60 stamps while Freddy has 5*20=100. Together they have 160
4 0
2 years ago
Can someone please explain y=mx+b
lana [24]

y=mx+b is slope-intercept form, which is a way of writing an equation for a line.

In this equation, m represents the slope and b represents the y-intercept.

x and y are the coordinates of a point on the resulting line.

Here are some examples:

y=3x+2

y=-2x-4

y=8x-1

y=-7x+5

7 0
3 years ago
To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil
Fiesta28 [93]

Answer:

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

H_0: The true average penetration is 50 mils

H_a: The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the <em>level of confidence</em> is α = 0.05 then the z_\alpha score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)

The z-score associated with this test is

z=\frac{\bar x-\mu}{s/\sqrt{n}}

where

\bar x = <em>mean of the sample</em>

\mu = <em>average established by the specification</em>

s = <em>standard deviation of the sample</em>

n = <em>size of the sample</em>

Computing this value of z we get z = 3.42

Since z >z_\alpha we can conclude that the sample has not met the required specification.

5 0
3 years ago
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