(5,0) and (-10-5) equation

Solve: the sum of two numbers is 70. one number is 4 times the other. what are the two numbers?

Thepotemich [5.8K]

Lets say two numbers are x and y

x+y=70-----------------------eq1

x=4y--------------------eq2

4y+y=70
5y=70
y=70/5
y=14

then x=4*14

x=56

8 0

3 years ago

Help me pls answer this

Rudik [331]

Answer:

right

Step-by-step explanation:

32/8=4

(m^5)(m^-3)=m^2

4 0

4 years ago

To join a health club, a person must pay a $75 one-time fee plus a charge of $30 per month. This situation can be presented by a

Rudiy27

Answer:

y = 30x + 75

Step-by-step explanation:

every month you're paying 30 dollars for a fee.

the 75 dollars is simply the "down payment" so to say.

m is the constant growth or the continuous growth or the charge per month.

5 0

3 years ago

Anne has a 10 3/4 pound bag of flour. Each time she bakes a batch of rolls she uses 2 1/6 pounds of flour. How many pounds of fl

LUCKY_DIMON [66]

She will have four pound left because 2 times 3 is 6 and 10 minus 6 is 4

3 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) $COV = 0.007255$

(c) $P(X > 980) = 0.00001\\\\$

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

4 years ago