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Gekata [30.6K]
3 years ago
10

Isabella deposited $2,000 in an account that earns 5% simple annual interest. After one year, how much interest will her investm

ent earns?
Mathematics
2 answers:
miskamm [114]3 years ago
8 0
Ans = $2,100
2000 x 1.05= 2100
1.00 = 100%
1.05 = 100% +5%
lorasvet [3.4K]3 years ago
7 0

Answer:

2100

Step-by-step explanation:

5% of 2000 = 2000 x 1/20 = 100.

2000+100=2100

Also, if it said how much interest will her investment PAY, then it would be $100 because of the 5% simple interest.

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What is the solution to the equation log(2x + 4) = 2? Round to the nearest thousandth, if necessary.
Kazeer [188]

Answer: 48

Step-by-step explanation:

log_{b}M=n--> M=b^{n} \\log(2x+4)=2\\b=10, M=2x+4, n= 2 \\\\2x+4=10^{2} \\2x+4=100\\2x=96\\x=48

5 0
3 years ago
I need to divide this equation:
Luba_88 [7]
1+7/11 = 11/11+7/11 = 18/11 2+1/2 = 4/2+1/2 = 5/2 (18/11)/(5/2) = (18/11)*(2/5) = 36/55
5 0
4 years ago
The game. Their scores are shoun below.
Harrizon [31]

Answer:

Barb is correct she won by 10 points

Darius's=480

Barb=490

5 0
3 years ago
You have 55,000 in your savings account that pays 2.5% annual interest and the inflation rate is 3.4%. How much buying power wil
diamong [38]
We solve the question as follows:
Simple interest=Principle×Rate×Time
Thus given:
p=$55000, R=2.5%, time= 1 year
thus
Interest=55000×0.025×1=$1375

To evaluate the amount required to keep up with the inflation, your interest rate should match the inflation rate otherwise prices are going up faster than the savings.
Required interest rate=55000×0.034×1=$1870

The buying power lost will be the difference between your required interest and actual interest.

Thus:
Buying power lost=1870-1375=$495

8 0
3 years ago
In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t
Angelina_Jolie [31]

Answer:

P(X = 0) = 0.027

P(X = 1) = 0.189

P(X = 2) = 0.441

P(X = 3)= 0.343

Step-by-step explanation:

The probability mass function P(X = x) is the probability that X happens x times.

When n trials happen, for each x \leq n, the probability mass function is given by:

P(X = x) = pe_{n,x}.p^{x}.(1-p)^{n-x}

In which p is the probability that the event happens.

pe_{n,x}, is the permutation of n elements with x repetitions(when there are multiple events happening(like one passes and two not passing)). It can be calculated by the following formula:

pe_{n,x} = \frac{n!}{x!}

The sum of all P(X=x) must be 1.

In this problem

We have 3 trials, so n = 3

The probability that a wafer pass a test is 0.7, so p = 0.7

Determine the probability mass function of the number of wafers from a lot that pass the test.

P(X = 0) = (0.7)^{0}.(0.3)^{3} = 0.027

P(X = 1) = pe_{3,1}.(0.7).(0.3)^{2} = 0.189

P(X = 2) = pe_{3,2}.(0.7)^{2}.(0.3) = 0.441

P(X = 3) = (0.7)^{3} = 0.343

5 0
4 years ago
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