All categories

3 years ago

Dan says he ran twice as far as Karim what possiblities for the distance they each could run

Mathematics

1 answer:

Lelechka [254]3 years ago

5 0

If Karim ran one mile, Dan ran two miles.
If Karim ran three miles, Dan ran six miles.

You might be interested in

The linear equation 3x – 11y = 10 has

storchak [24]

Answer:

Correct option: c) infinitely many solutions

Step-by-step explanation:

We have an equation with two variables, so as we have a system with less equations (just one) than variables (two), we have an infinite number of solutions.

To find some solutions, we can choose values for x and then just calculate the value of y:

x = 0:

-11y = 10 -> y = -10/11

x = 1:

3 - 11y = 10 -> y = -7/11

x = 2:

6 - 11y = 10 -> y = -4/11

And so on.

So the correct option is c): infinitely many solutions

7 0

3 years ago

I got more points so here is more

BartSMP [9]

Answer:

woah thankssss a lotttt

7 0

3 years ago

Read 2 more answers

Can someone please help me with these questions?

tensa zangetsu [6.8K]

Answer:

c and a i think

Step-by-step explanation:

7 0

2 years ago

An artist creates a cone-shaped sculpture for an art exhibit. If the sculpture is 55 feet tall and has a base with a circumfere

yulyashka [42]

Answer:

4164.69504π

Step-by-step explanation:

So the volume of a cone is πr^2 * h / 3

r = 15.072 / π

h = 55

15.072 / π squared * π = 227.165184π

227.165184π * 55 / 3 = 4164.69504π

4 0

3 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

3 years ago