What is the smallest mumber to be added to 11650 so that the sum will be a perfect square

A bank offers a savings account with a yearly simple interest rate of 2%. Suppose you deposit $550 into a savings account. How m

Bingel [31]

Answer:

principal, or amount your account starts with. This could be a starting investment, or the starting amount of a loan. Interest, in its most simple form, is calculated as a percent of the principal. For example, if you borrowed $100 from a friend and agree to repay it with 5% interest, then the amount of interest you would pay would just be 5% of 100: $100(0.05) = $5. The total amount you would repay would be $105, the original principal plus the interest.

7 0

3 years ago

Line A is

kirill [66]

Step-by-step explanation:

the answer is a bc it just makes sense

3 0

1 year ago

Read 2 more answers

Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and relea

Talja [164]

Answer:

a) For this case the random variable X follows a hypergometric distribution.

b) $E(X)= n\frac{M}{N}=10 \frac{5}{25}=2$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1$

c) $P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565$

d) $P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474$

Step-by-step explanation:

The hypergometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

$P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}$

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes

The expected value and variance for this distribution are given by:

$E(X)= n\frac{M}{N}$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}$

a. What is the distribution of X?

For this case the random variable X follows a hypergometric distribution.

b. Compute the values for E(X) and Var(X)

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

$E(X)= n\frac{M}{N}=10 \frac{5}{25}=2$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1$

c. What is the probability that none of the animals in the second sample are tagged?

So for this case we want this probability:

$P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565$

d. What is the probability that all of the animals in the second sample are tagged?

So for this case we want this probability:

$P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474$

4 0

3 years ago

A batch of 40 parts contains six defects. If two parts are drawn randomly one at a time without replacement, what is the probabi

adell [148]

Answer:

a) Therefore, the probability is P=1/52.

b) Therefore, the probability is P=9/400.

Step-by-step explanation:

We know that a batch of 40 parts contains six defects.

a) We calculate the probability that the both parts are defective, if two parts are drawn randomly one at a time without replacement.

The probability for the first is 6/40.

The probability for the second is 5/39.

We get:

$P=\frac{6}{40}\cdot \frac{5}{39}\\\\P=\frac{1}{52}$

Therefore, the probability is P=1/52.

b) We calculate the probability that the both parts are defective, if two parts are drawn randomly one at a time with replacement.

The probability for the first is 6/40.

The probability for the second is 6/40.

We get:

$P=\frac{6}{40}\cdot \frac{6}{40}\\\\P=\frac{9}{400}$

Therefore, the probability is P=9/400.

6 0

3 years ago

Please please help! I'm stuck on this

Phantasy [73]

Answer:

A'(-4 , 10)

B'(-4, 12)

C'(0, 12)

D'(0, 10)

Step-by-step explanation:

Given in the question are 4 coordinates of Quadrilateral ABCD

A(−2, 2)

B(−2, 4)

C(2, 4)

D(2, 2)

The rule is (x − 2, y + 8), that means

the translation is 2 units to the left and 8 units up

Applying the rule of the translation

Step1

A(-2, 2) = A'(-2-2 , 2+8) = A'(-4 , 10)

Step 2

B(−2, 4) = B'(-2-2 , 4+8) = B'(-4, 12)

Step 3

C(2, 4) = C'(2-2 , 4+8) = C'(0, 12)

Step 4

D(2, 2) = D'(2-2 , 2+8) = D'(0, 10)

Describe what characteristics you would find if the corresponding vertices were connected with line segments.

If point A was connected to point A', point B was connected to point B',point C was connected to point C', point D was connected to point D' then all of these line will be parallel to each other, since they were all transformed by the same amount (x-2, y+8).

5 0

3 years ago

What is the smallest mumber to be added to 11650 so that the sum will be a perfect square​

What is the smallest mumber to be added to 11650 so that the sum will be a perfect square