Solve the following equation.

Without multiplying, is the product of 45x 10 or 50x10 greater? Explain

Aleonysh [2.5K]

Answer:

50x10 is greater

Step-by-step explanation:

Multiplying by ten is basically adding a zero at the end of the number. So if you add a zero at the end of both the numbers you see that 50x10 would be greater.

Or you can look at the numbers and see that 50x10 has bigger numbers than 45x10 so 50x10 is greater

4 0

3 years ago

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Whats the definition of a circle?

Katarina [22]

Hello!

A circle is a round plane figure what's boundary is made of points from the center..

Hope this helps! ☺♥

6 0

3 years ago

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Find the surface area of the pyramid.

zubka84 [21]

Answer: 280.8

Step-by-step explanation:

3 0

3 years ago

You are setting up a zip line in your yard. You map out your yard in a coordinate plane. An equation of the line representing th

neonofarm [45]

Answer:

Distance is 6.7units to the nearest tenth.

Explanation:

To solve this problem, we the distance formula. And this is given as:

$D = \frac{|a x + b y + c|}{\sqrt{a^2 + b^2}}$

We were given the equation:

y = 2 x + 4

This can also be rewritten as:

y – 2 x – 4 = 0.

Therefore, a = -2, b = 1 and c = -4

Also, we were given the points:

(x, y) = (–4, 11)

Making use of the above formula for distance, we have:

$D = \frac{|-2\times -4 + 1\times 11 + -4|}{\sqrt{-2^2 + -4^2}}$

$D = \frac{15}{\sqrt{5}}$

D = 6.7units (to the nearest tenth)

8 0

3 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago