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3 years ago

5

Identify how the Virginia declaration of rights helped lead to an expansion of civil rights both in the state of Virginia and la

ter in the entire United States

1 answer:

-BARSIC- [3]3 years ago

7 0

Answer:

yeah

Step-by-step explanation:

wrong subject dude

things go wrong sometimes

you lost 46 points

$\lim_{n \to \infty} a_n$

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Spencer took $21 with him to the mall. At the music store he spent 2/3 of his money on a DVD. Then spent 1/4 on a drink. How muc

frutty [35]

After the music store he had $14 left, because 21:3=7 and 7x2 equals 14.

After the drink, he has $3.5 left, because 14:4=3.5 and 3.5x1=3.5.

Conclusion: that is one expensive drink. (Answer is 3 dollars and 50 cents. Hope this helped:)

6 0

3 years ago

Read 2 more answers

Tayâ€“Sachs Disease Tayâ€“Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carri

BARSIC [14]

Answer:

a) 0.0156 = 1.56% probability that all children will develop the disease.

b) 0.4219 = 42.19% probability that only one child will develop the disease.

c) 0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that their offspring will develop the disease is approximately .25.

This means that $p = 0.25$

Three children:

This means that $n = 3$

Question a:

This is P(X = 3). So

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

$P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156$

0.0156 = 1.56% probability that all children will develop the disease.

Question b:

This is P(X = 1). So

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

$P(X = 1) = C_{3,1}.(0.25)^{1}.(0.75)^{2} = 0.4219$

0.4219 = 42.19% probability that only one child will develop the disease.

c. The third child will develop Tayâ€“Sachs disease, given that the first two did not.

Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

Third develops, which 0.25 probability. So

$p = 0.75*0.75*0.25 = 0.1406$

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

6 0

3 years ago

The mean weight of the domesticated house cat is 9.2 pounds with a standard deviation of 3.4 pounds. If a cat has a z-score of -

ra1l [238]

Answer:

4.78 pounds

Step-by-step explanation:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 9.2 pound

σ is the population standard deviation = 3.4 pounds

z = 1.3

We are to find x

Hence,

-1.3 = x - 9.2/3.4

Cross Multiply

-1.3 × 3.4 = x - 9.2

-4.42 = x - 9.2

x = -4.42 + 9.2

x = 4.78 pounds.

Therefore the weight of the cat is 4.78 pounds

5 0

3 years ago

Pls explain to me how to find the value of x.

sladkih [1.3K]

Here are the pictures of the steps I took when solving it as well as the answer to what X is

4 0

3 years ago

Steps factoring polynomials using gcf for 9x²+9x

Molodets [167]

Answer:

Step-by-step explanation:

Step 1: Identify the GCF of the polynomial.

Step 2: Divide the GCF out of every term of the polynomial. ...

Step 1: Identify the GCF of the polynomial. ...

Step 2: Divide the GCF out of every term of the polynomial.

Step 1: Identify the GCF of the polynomial. ...

Step 2: Divide the GCF out of every term of the polynomial .

6 0

3 years ago