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vfiekz [6]
3 years ago
10

(4x-1)(4x+1)-3=15x^2

Mathematics
1 answer:
algol133 years ago
8 0
(16x^2+4x-4x-1)-3=15x^2
16x^2-1-3=15x^2
16x^2-4=15x^2
X^2=4
X=+or-2
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The set of ordered pairs of the form (x,y) shown below represent points on a graph for a direct variation.
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The answer is B y=2/3x
6 0
3 years ago
A right triangle has an angle that measures 34 and the adjacent side measure 17. what is the lenght of the hypotenuse to the nea
olga2289 [7]
The length of the hypotenuse is 38
7 0
3 years ago
From a random sample of 58 businesses, it is found that the mean time the owner spends on administrative issues each week is 21.
zzz [600]

Answer: (20.86, 22.52)

Step-by-step explanation:

Formula to find the confidence interval for population mean :-

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}

, where \overline{x} = sample mean.

z*= critical z-value

n= sample size.

\sigma = Population standard deviation.

By considering the given question , we have

\overline{x}= 21.69

\sigma=3.23

n= 58

Using z-table, the critical z-value for 95% confidence = z* = 1.96

Then, 95% confidence interval for the amount of time spent on administrative issues will be :

21.69\pm (1.96)\dfrac{3.23}{\sqrt{58}}

=21.69\pm (1.96)\dfrac{1.7}{7.61577}

=21.69\pm (1.96)(0.223221)

\approx21.69\pm0.83

=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)

Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)

7 0
3 years ago
Use technology and the given confidence level and sample data to find the confidence interval for the population mean mu . assum
attashe74 [19]

To solve for the confidence interval for the population mean mu, we can use the formula:

Confidence interval = x ± z * s / sqrt (n)

where x is the sample mean, s is the standard deviation, and n is the sample size

 

At 95% confidence level, the value of z is equivalent to:

z = 1.96

 

Therefore substituting the given values into the equation:

Confidence interval = 3 ± 1.96 * 5.8 / sqrt (51)

Confidence interval = 3 ± 1.59

Confidence interval = 1.41, 4.59

 

Therefore the population mean mu has an approximate range or confidence interval from 1.41 kg to 4.59 kg.

6 0
3 years ago
Here are summary statistics for randomly selected weights of newborn​ girls: nequals202​, x overbarequals28.3 ​hg, sequals6.1 hg
Lorico [155]

Answer:

The confidence interval is 27.5 hg less than mu less than 29.1 hg

(A) Yes, because the confidence interval limits are not similar.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 28.3 hg

sd = 6.1 hg

n = 202

degree of freedom = n-1 = 202-1 = 201

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196

E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg

Lower limit = mean - E = 28.3 0.8 = 27.5 hg

Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg

95% confidence interval is (27.5, 29.1)

When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17

7 0
3 years ago
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