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Sav [38]
3 years ago
7

What’s the answer !?

Mathematics
1 answer:
svetoff [14.1K]3 years ago
5 0

Answer:

8

Step-by-step explanation:

16(³/⁴) =

= (2⁴)(³/⁴)

= 2⁴ˣ³/⁴

= 2³

= 8

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For what values of the variable, do the following fractions exist: y^2-1/y+y/y-3
ale4655 [162]

Answer:

Remember that the division by zero is not defined, this is the criteria that we will use in this case.

1) \frac{y^2 - 1}{y}  + \frac{y}{y - 3}

So the fractions are defined such that the denominator is never zero.

For the first fraction, the denominator is zero when y = 0

and for the second fraction, the denominator is zero when y = 3

Then the fractions exist for all real values except for y = 0 or y = 3

we can write this as:

R / {0} U { 3}

(the set of all real numbers except the elements 0 and 3)

2) \frac{b + 4}{b^2 + 7}

Let's see the values of b such that the denominator is zero:

b^2 + 7 = 0

b^2 = -7

b = √-7

This is a complex value, assuming that b can only be a real number, there is no value of b such that the denominator is zero, then the fraction is defined for every real number.

The allowed values are R, the set of all real numbers.

3) \frac{a}{a*(a - 1) - 1}

Again, we need to find the value of a such that the denominator is zero.

a*(a - 1) - 1 = a^2 - a - 1

So we need to solve:

a^2 - a - 1 = 0

We can use the Bhaskara's formula, the two values of a are given by:

a = \frac{-(-1) \pm \sqrt{(-1)^2 + 4*1*(-1)}  }{2*1}  = \frac{1 \pm \sqrt{5} }{2}

Then the two values of a that are not allowed are:

a = (1 + √5)/2

a = (1 - √5)/2

Then the allowed values of a are:

R / {(1 + √5)/2} U {(1 - √5)/2}

7 0
2 years ago
Suppose a sample of 972972 tenth graders is drawn. Of the students sampled, 700700 read above the eighth grade level. Using the
max2010maxim [7]

Answer:

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

972 students, 700 read above the eight grade level. We want the confidence interver for the proportion of those who read at or below the 8th grade level. 972 - 700 = 272, so n = 972, \pi = \frac{272}{972} = 0.28

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.259

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 + 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.301

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

3 0
3 years ago
Which statement is correct?
VladimirAG [237]
C is the correct answer tell me if its correct
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45 centimeters is equivalent to how many inches?
Elza [17]
It’s about 17.7 inches
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Ivanshal [37]
480? if i’m doing it right
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3 years ago
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