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2 years ago

X/z+y^2-x evaluate where x=-2 y=4 and z=16

Mathematics

1 answer:

lozanna [386]2 years ago

4 0

Answer:

Solution

X=-2

y=4

z=16

then,

-2/16+16×16--2

=257.875

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mike says 3/3 of his fraction model is shaded blue. Ryan says 6/6 of the same model is shaded blue . are the two fractions equiv

Jet001 [13]

Yes, they are equivalent. Another equivalent fraction would be any fraction that equals to 1. Eg. 9/9

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Please answer fasttt

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Answer:

7.5 lb/cu ft

Step-by-step explanation:

75 lb / 10 cu ft = 7.5 lb/cu ft

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2 years ago

A b c or d. use pic to solve

Neporo4naja [7]

Answer:

In Heather's solution to the problem, she wrote and solved an equation.

Her work is:

Step 1: 1.08(x +9.01 +0.98 +5.01) = 21.87

Solving like terms

1.08(x+15}=21.87

dividing both side by 1.08

we get

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subtracting both side by 13.9198

x=20.25-15

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According to correction

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Option 1st

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2 years ago

Answer the question on the picture

goblinko [34]

Answer:

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hope it helps.

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3 years ago

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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:

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

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

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

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.

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3 years ago