Ask question

Ask question

All categories

Anestetic [448]

3 years ago

14

Diana invested $3000 in a savings account for 3 years. She earned $450 in interest over that time period. What interest rate did

she earn?

2 answers:

Aleks04 [339]3 years ago

7 0

So she invested $3000. 450 of 3000 would be 0.15 which is 15%. So in one year, interest rate would be 15% divided by 3 years = 5 %
Best of luck!

REY [17]3 years ago

3 0

formula I=Prt 450= (3000) r (3) 450=9000r 450/9000=9000r 0.05=r

You might be interested in

What expressions are equivalent to 4(4a+5)

Firlakuza [10]

Answer:

<h3>16a + 20</h3>

Step-by-step explanation:

4(4a + 5) <em>use distributive property</em>

= (4)(4a) + (4)(5) = 16a + 20

5 0

3 years ago

What is 1-cos(6x)=___

FromTheMoon [43]

Answer:

2sin^2(3x)

Step-by-step explanation:

3 0

2 years ago

Please help!! i’ll mark as brainliest!!

liraira [26]

The answer should be 4/3

8 0

3 years ago

Randy and Susie Moser are newlyweds. Randy drives their car to his job, twenty-seven miles away from home, each day. If Randy an

Andrews [41]

The answer is $834.10

3 0

3 years ago

Read 2 more answers

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

3 years ago