Please someone come thru for me idk how to get these answers

16 of the 22 students in Mrs. Burnham's class passed a quiz. About what

ser-zykov [4K]

Answer:

72.73

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

An aptitude test has a mean score of 80 and a standard deviation of 5. The population of scores is normally distributed. What pr

konstantin123 [22]

Answer:

2.28% of tests has scores over 90.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 80, \sigma = 5$

What proportion of tests has scores over 90?

This proportion is 1 subtracted by the pvalue of Z when X = 90. So

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

$Z = \frac{90 - 80}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

So 1-0.9772 = 0.0228 = 2.28% of tests has scores over 90.

8 0

2 years ago

solve for each please i really need help if u want to help me with my test and i get an a or a b i will give u 500 dollars

Len [333]

Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

Every x has only one value of y

x appears once in ordered pairs

<u><em>Examples:</em></u>

The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)

The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)

The domain is the set of values of x

The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

3 0

2 years ago

(10 pts) (a) (2 pts) What is the difference between an ordinary differential equation and an initial value problem? (b) (2 pts)

laiz [17]

Answer:

Step-by-step explanation:

(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.

(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.

(C) Example of a second order linear ODE:

M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)

The equation will be homogeneous if K(t)=0 and heterogeneous if $K(t)\neq 0$