<img src="https://tex.z-dn.net/?f=10%20%5Cdiv%20%20%5Csqrt%7B6x%7D%20" id="TexFormula1" title="10 \div \sqrt{6x} " alt="10 \div

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scZoUnD [109]

2 years ago

\sqrt{6x} " align="absmiddle" class="latex-formula">

Mathematics

2 answers:

Aliun [14]2 years ago

8 0

The answer is what the guy below me said

Oxana [17]2 years ago

3 0

Unfortunately, you haven't shared the instructions. My guess is that you were supposed to "rationalize the denominator," that is, remove the radical from the denominator.

10 sqrt(6x) 10sqrt(6x)
------------ * ------------- = ----------------
sqrt(6x) sqrt(6x) 6x

This is the correct form of the answer; it has no radical in the denominator.

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Which of the following statements is true about the quadratic function f(x) = ax2 + bx + c?

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Step-by-step explanation:

A. The constants a, b, and c must be real numbers with a not ever equal to zero.

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1 year ago

Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask

Kay [80]

Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

$P(x=r)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

$P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}$

$=56(0.4)^5 (0.6)^3$

$=0.12386304$

(b) the probability of the number that say they would feel secure is more than five

$P(X>5) = P(X=6)+ P(X=7) + P(X=8)$

$=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}$

$=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8$

$=0.04980736$

(c) the probability of the number that say they would feel secure is at most five

$P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$

$=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}$