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Veseljchak [2.6K]

3 years ago

7

Could you help me with this equation:

%20%3D%200" id="TexFormula1" title="x^{3} + 5x^{2} - 18x + 3 = 0" alt="x^{3} + 5x^{2} - 18x + 3 = 0" align="absmiddle" class="latex-formula">

1 answer:

lord [1]3 years ago

3 0

if you want to solve it without a graph you're going to need to use the cubic formula, I'm unsure how else to solve it

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Kendra received a bonus that was 30% of her monthly earnings. If her monthly earnings were $930, how much was Kendra's bonus?

Lady bird [3.3K]

Answer:

$279

Step-by-step explanation:

100% = $930

30% = ???

(30/100) × 930

=$279

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2 years ago

What is 2/8 x 4/5<br> a.6/40<br> b.8/40<br> c.6/13<br> d.8/13

Leya [2.2K]

The answer is B 8/40

5 0

3 years ago

Read 2 more answers

lys-0071 [83]

Answer:

f(x)+g(x)=4x^3+x−7

Step-by-step explanation:

To find f(x) + g(x) add the expressions together by combining like terms. Recall like terms are terms with the same base and exponent.

f(x) + g(x) = x - 7 + 4x^3

f(x) + g(x) = 4x^3 + x -7

6 0

4 years ago

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

$P(X \leq 1) + P(X > 1) = 1$

$P(X > 1) = 1 - P(X \leq 1)$

We also have that:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

So

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

3 0

3 years ago

Consider this right triangle.

andrew-mc [135]

Answer:

5/13

Step-by-step explanation:

4 0

3 years ago