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Alisiya [41]
3 years ago
11

I need help pls is due in 40 minutes

Mathematics
1 answer:
liubo4ka [24]3 years ago
6 0

Answer:

is this due already

Step-by-step explanation:

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A student solved log4(2x – 12) = 3, as shown. Which did you include in your answer?
siniylev [52]

Answer:

x = 38

Step-by-step explanation:

using the law of logarithms

• log_{b} x = n ⇔ x = b^{n}, hence

log_{4}(2x - 12) = 3 ⇒ 2x - 12 =  4^{3} = 64

add 12 to both sides

2x = 76 ( divide both sides by 2 )

x = 38


5 0
4 years ago
4 to the 4th power equals 256. Explain how to use that fact to more quickly evaluate 4 to the 5th power.
Ray Of Light [21]

Answer: Because 4 is the base of what is being exponentially multiplied, you can multiply 256 by 4 to get 4^5

6 0
3 years ago
Read 2 more answers
Answer for brainlest for 25 points
Gnom [1K]

Answer:

1. 23/4  ,  2. 5.555...   , 3. 5 1/2   ,    4. √28

you want to make them all into decimals so you can compare then easier

23/4 would be 5.75

√28 (just put it in a calculator)  would be 5.29

5 1/2  would be 5.5  (make  it into a improper fraction 11/2 then do 11/2 and you get 5.5)

and you have 5.555.

so...

5.75

5.29

5.5

5.55555

first is 5.75 it's bigger because of the .(75)

then 5.555 because it's .(55)

then 5.5 because it's .(50)

and the last one 5.(29)

7 0
3 years ago
A. Student GPAs: Bob’s z-score z = + 1.71 μ = 2.98 σ = 0.36 b. Weekly work hours: Sarah’s z-score z = + 1.18 μ = 21.6 σ = 7.1 c.
hjlf

Answer:

a) x = \mu +z*\sigma

And replacing we got:

x= 2.98 + 1.71*0.36 = 3.5956

b) x = \mu +z*\sigma

And replacing we got:

x= 21.6 + 1.18*7.1 = 29.978

c) x = \mu -z*\sigma

And replacing we got:

x= 150 - 1.35*40= 96

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable of interest for this case. We define the z score with the following formula:

z=\frac{x-\mu}{\sigma}

And for this case we know that z = 1.71, \mu = 2.98,\sigma = 0.36

If we solve for x from the z score formula we got:

x = \mu +z*\sigma

And replacing we got:

x= 2.98 + 1.71*0.36 = 3.5956

Part b

Let X the random variable of interest for this case. We define the z score with the following formula:

z=\frac{x-\mu}{\sigma}

And for this case we know that z = 1.18, \mu = 21.6,\sigma = 7.1

If we solve for x from the z score formula we got:

x = \mu +z*\sigma

And replacing we got:

x= 21.6 + 1.18*7.1 = 29.978

Part c

Let X the random variable of interest for this case. We define the z score with the following formula:

z=\frac{x-\mu}{\sigma}

And for this case we know that z = -1.35, \mu = 150,\sigma = 40

If we solve for x from the z score formula we got:

x = \mu -z*\sigma

And replacing we got:

x= 150 - 1.35*40= 96

5 0
3 years ago
A random sample of 225 individuals working in a large city indicated that 45 are dissatisfied with their working conditions. Bas
Lorico [155]

Answer:

CI 90 %  =  [  0,165  ;  0,235 ]

Lower limit     0,165

upper limit      0,235

Step-by-step explanation:

Sample information:

sample size    n  =  225

number of dissatisfied  individuals   x  =  45

p = 45/225

p = 0,2    and  q =  1 -  p   q  =  1  -  0,2   q  =  0,8

p*n = 0,2*225   =  45    and   q*n   =  0,8*225  =  180

p*n  and q*n  big enough to use the approximation of binomial distribution to normal distribution

90 % of Confidence Interval   then a significance level is  α  = 10%

α  =  0,1 in z table we get z(c) for that significance level

z(c) = 1,28

CI 90 %   =  [  p  ±  z(c)*√(p*q)/n ]

z(c) * √(p*q)/n    =  1,28 *  √ ( 0,2*0,8)/225

z(c) * √(p*q)/n    =  1,28 * 0,027

z(c) * √(p*q)/n    =  0,035

CI 90 %  =  [  0,2  ±  0,035 ]

CI 90 %  =  [  0,165  ;  0,235 ]

Lower limit     0,165

upper limit      0,235

3 0
3 years ago
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