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3 years ago

If 3£A and 3€B, then which of the following statements is not true?

Mathematics

1 answer:

WITCHER [35]3 years ago

8 0

The answer is C. 3 is an element of A n B

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50 laps in 10 day = how many laps in 1 day?

Natali [406]

5 laps in 1 day

50/10 = 5

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3 years ago

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Which exponential function is represented by the values in the table?

svlad2 [7]

Here we use the equation

$y = a(b)^x$

Taking points (0,4), (1,2)

Substituting the point (0,4) , we will get

$4 = a(b)^0
\\
4 = a(1) \\
a =4$

Substituting (1,2) we will get

$2 = 4 (b)^1
\\
2/4 = b
\\
b = 1/2$

So we have

$a = 4, b = 1/2$

Therefore , required equation is

$y = 4(1/2)^x$

7 0

3 years ago

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Use the graph to evaluate the function. f(-5) = y -8 -6

DENIUS [597]

Answer:

-4

Step-by-step explanation:

Go to x value of -5 and go down until you touch line.

3 0

3 years ago

#17-1: Simplify this complex fraction. 1/3 / 6/9

Gnesinka [82]

$\huge{\boxed{\frac{1}{2}}}$

To divide by a fraction, you multiply by its reciprocal, which you find by flipping the numerator and the denominator. $\frac{1}{3} \div \frac{6}{9} = \frac{1}{3} * \frac{9}{6}$

Multiply the numerators and the denominators separately. $\frac{1}{3} * \frac{9}{6} = \frac{1*9}{3*6} = \frac{9}{18}$

Simplify by dividing both sides of the equation by $9$ . $\frac{9 \div 9}{18 \div 9} = \boxed{\frac{1}{2}}$

3 0

3 years ago

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A political candidate has asked you to conduct a poll to determine what percentage of people support him. If the candidate only

Nadusha1986 [10]

Answer:

The sample size required is, n = 502.

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$

The margin of error is:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p\ \cdot (1-\hat p)}{n}}$

Assume that 50% of the people would support this political candidate.

The margin of error is, MOE = 0.05.

The critical value of z for 97.5% confidence level is:

z = 2.24

Compute the sample size as follows:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p\ \cdot (1-\hat p)}{n}}$

$n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}$

$=[\frac{2.24\times \sqrt{0.50(1-0.50)}}{0.05}]^{2}\\\\=501.76\\\\\approx 502$

Thus, the sample size required is, n = 502.

4 0

3 years ago