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inessss [21]
3 years ago
15

im rushing and have like 20 more things so if someone could actually help me right answers, please. T~T

Mathematics
2 answers:
igomit [66]3 years ago
7 0
1. a. 0=0, 4x=4x, -18=-18
b. 3
2.a. 0=-2, 2= 0
b. 2
Sphinxa [80]3 years ago
5 0

Answer:

1

Step-by-step explanation:

4x-18=x-18=3x

the two 18 cancel out

4x=x+3x

combine like terms

4x=4x

dvide and any number divided by its self is 1

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Will factor the express 36xyz + 24xy − 16x by dividing each term by a common fact
sesenic [268]

\\ \bull\tt\longmapsto 36xyz+24xy-16x

  • Divide by 4

\\ \bull\tt\longmapsto \dfrac{36xyz}{4}+\dfrac{24xy}{4}-\dfrac{16x}{4}

\\ \bull\tt\longmapsto 9xyz+6xy-4x

  • Divide by x

\\ \bull\tt\longmapsto \dfrac{9xyz}{x}+\dfrac{6xy}{x}-\dfrac{4x}{x}

\\ \bull\tt\longmapsto 9yz+6y-4

6 0
3 years ago
Read 2 more answers
I'll give you the brainliest
Helen [10]

Answer:

>

Step-by-step explanation:

sqrt{17} > sqrt{11}

4 0
3 years ago
Read 2 more answers
The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However,
horsena [70]

Answer:

The mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

Step-by-step explanation:

We are given that records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.

On a typical​ Saturday, the​ oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Assuming that data follows normal distribution.

Let \bar X = <u><em>sample mean time</em></u>

The z score probability distribution for sample mean is given by;

                          Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 16.2 minutes

            \sigma = standard deviation = 3.4 minutes

            n = sample size = 35 oil changes

<u>Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;</u>

<u></u>

          P(X \leq x) = 0.10          {where x is required mean oil-change time}

          P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

          P(Z \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

Now, in the z table the critical value of X which represents the below 10% of the probability area is given as -1.282, that means;

                  \frac{x-16.2}{\frac{3.4}{\sqrt{35} } }  =  -1.282

               x - 16.2  =  -1.282 \times {\frac{3.4}{\sqrt{35} } }

                         x  =  16.2 - 0.74 = <u>15.46 minutes</u>

Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

8 0
3 years ago
Fredrick threw a ball from the top of a building. The height of the ball, in feet, after it is thrown is modeled by the function
V125BC [204]

Answer:

  • E. 5.2 seconds

Step-by-step explanation:

Need to find the value of t, when h(t) = 0

  • -16t^2 +64t + 100 = 0
  • 4t^2 - 16t - 25 = 0
  • 4t^2 - 2*2t*4 + 16 = 41
  • (2t - 4)^2 = 41
  • 2t - 4 = √41
  • 2t - 4 = 6.4
  • 2t = 10.4
  • t = 5.2

Answer is 5.2 seconds

4 0
3 years ago
What is the middle value of the following data set? 33, 36, 24 31, 39,44,32
Sonja [21]
Mean ( average) = 34.14286
Median (middle ) = 33
Hope that helped or the answer u was looking for
6 0
3 years ago
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