Fred lives on the 16th floor

Can y’all also show me how to do this so I won’t have to come back here for help plz

Igoryamba

Answer:

the value of x = 56

Step-by-step explanation:

if both are similar figures, then ratio of their

sides will be constant , so

=》14 / 17 = x / 68

=》14 / 17 × 68 = x

=》14 × 4 = x

=》x = 56

hence, value of x = 56.

i hope you got it.

8 0

2 years ago

A central angle of a circle measures 42 degrees. What else can be shown to measure 42 degrees?

Slav-nsk [51]

Answer:

Vertically opposite angles

Step-by-step explanation:

Given

$\theta = 42^\circ$

Required

What else can measure $\theta = 42^\circ$

The question has missing options. So, I will give a general explanation.

Using the attached image as a point of reference.

Given that:

$\theta = 42^\circ$

The value of b is calculated as:

$b = \theta$

Because b and $\theta$ are vertically opposite angles

This gives:

$b = 42^\circ$

So, one possible answer to the question is a vertically opposite angle

4 0

3 years ago

William gave 4 football cards to each

andreev551 [17]

Answer:

Cannot be determined

Step-by-step explanation:

There is not enough information to solve this question. You would need to know how many friends he has in order to solve this.

4 0

2 years ago

Find the solution of the system x′=−4y,y′=−2x, where primes indicate derivatives with respect to t, that satisfies the initial c

Andreas93 [3]

Answer: $2y^2 = x^2+4$

Step-by-step explanation:

given is a differntial equation in parametric form as

$x'=-4y\\y'=-2x$

We can find dy/dx by dividing y' by x'

$\frac{dy}{dx} =\frac{-2x}{-4y} \\4ydy =2x dx\\2y^2 = x^2+C$

(By separating the variables and integrating)

C is arbitrary constant

When x=-2, y =-2

Substitute to get

$8=4+C\\C=4$

So solutoin is

$2y^2 = x^2+4$

is the solution

5 0

3 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2500 individ

Ne4ueva [31]

Answer:

The approximate distribution of the number who carry this gene is approximately normal with mean $\mu = 5$ and standard deviation $\sigma = 2.23$

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they carry the defective gene that causes inherited colon cancer, or they do not. The probability of a person carrying this gene is independent from other people. So the binomial probability distribution is used to solve this question.

A sample of 2500 individuals is quite large, so we use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer.

This means that $p = \frac{1}{500} = 0.002$

In a sample of 2500 individuals, what is the approximate distribution of the number who carry this gene?

$n = 2500$

So

$\mu = E(X) = np = 2500*0.002 = 5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2500*0.002*0.998} = 2.23$

So the approximate distribution of the number who carry this gene is approximately normal with mean $\mu = 5$ and standard deviation $\sigma = 2.23$

4 0

3 years ago