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3 years ago

Round 6.73524 to the Nearest hundredth. 6.73 6.734 6.735 6.74

Mathematics

2 answers:

Alexandra [31]3 years ago

8 0

We can figure this out by looking at the number after the hundredths place:

6.73<u>5</u>24

If the number is 5 or above, you round the number in the hundredths place up. But if it is 4 or below, you leave the number the same.

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Hope this helps!! Good luck.

docker41 [41]3 years ago

5 0

Answer:

The answer is 6.74

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The measure of one of the angles formed by two parallel lines and a transversal is 45°. Is it possible for the measure of any of

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4 years ago

Which expressions are equivalent to 25x^4–64? select three options.

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<h2>The equivalent expression of 25 $x^4$ - 64 are "a), b) and c)".</h2>

Step-by-step explanation:

The given expression is:

25 $x^4$ - 64

To find, the equivalent expressions are:

a) 25 $x^4$ + 40x - 40x - 64

= 25 $x^4$ - 64, is the equivalent expression.

b) 25 $x^4$ + 13x - 13x - 64

= 25 $x^4$ - 64, is the equivalent expression.

c) (5 $x^2$ + 8)(5 $x^2$ - 8)

Using the algebraic identity,

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= $(5x^2)^2-8^2$

= 25 $x^4$ - 64, is the equivalent expression.

d) ( $x^2$ + 13)( $x^2$ - 13)

Using the algebraic identity,

(a + b)(a - b) = $a^{2}-b^{2}$

= $(x^2)^2-13^2$

= $x^4$ - 169, is not a equivalent expression.

e) $(5x^2-8)^2$

Using the algebraic identity,

$(a-b)^{2} =a^{2} -2ab+b^2$

$= (5x^2)^{2} -2(5x^2)(8)+8^2$

$= 25x^4 -80x^2+64$ , is not a equivalent expression.

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4 years ago

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Dominik [7]

Answer:

The probability that a person is audited more than twice is 0.0037 .

Step-by-step explanation:

We are given that The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year.

We are interested in the expected number of audits a person with that income has in a 16-year period

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So, We will use binomial distribution

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So, Probability of failure = q=1-0.02=0.98

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So, $P(x \geq 2)= 1-(P(x=0)+P(x=1)+P(x=2)$

Formula : $^nC_x p^x q^{n-x}$

So, $P(x \geq 2)= 1-(^{16}C_0 {(0.02)}^0 {(0.98)}^{16-0}+^{16}C_1 {(0.02)}^1 {(0.98)}^{16-1}+ ^{16}C_2 {(0.02)}^2 {(0.98)}^{16-2})$ $P(x \geq 2)= 1-(\frac{16!}{0!(16-0)!} {(0.02)}^0 {(0.98)}^{16-0}+\frac{16!}{1!(16-1)!} {(0.02)}^1 {(0.98)}^{16-1}+\frac{16!}{2!(16-2)!} {(0.02)}^2 {(0.98)}^{16-2})$

$P(x \geq 2)= 1-(0.9963)$

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