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3 years ago

9

A pair of sneakers is priced at $150 and is on sale. Yolanda pays $105 for a pair of the sneakers. What percent is the markdown

on the sneakers?

1 answer:

Otrada [13]3 years ago

7 0

638383873737377373893993

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Dr. Ashby is studying the heart rate of adult men. Preliminary data suggests that the standard deviation of heart rates for adul

jok3333 [9.3K]

Answer:

n=42 adult men

Step-by-step explanation:

-The minimum sample size to get a desired margin of error, E is calculated using the formula:

$n\geq( \frac{z\sigma}{E})^2$

-Given that $\sigma=7.5, \ \ E=3\ , \ z_{0.005}=2.58$ , we can substitute to solve for n in the above formula:

$n\geq (\frac{z\sigma}{E})^2\\\\\therefore n=(\frac{z_{\alpha/2}\sigma}{E})^2\\\\=(\frac{2.58\times7.5}{3})^2\\\\=41.6025\approx42$

Hence, the desired minimum sample size, n is 42 adult men.

4 0

3 years ago

Find the value of x in the isosceles triangle shown below

Vesnalui [34]

Answer:

the value of x is 8

x=8

HOPE IT'S HELP

6 0

3 years ago

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.

4 0

3 years ago

If \orange{\angle ADC}∠ADCstart color #ffa500, angle, A, D, C, end color #ffa500 measures 23^\circ23 ∘ 23, degrees, what does \b

3241004551 [841]

Can you just post the pic

5 0

2 years ago

Read 2 more answers

Two part-time instructors are hired by the Department of Statistics and each is assigned at random to teach a single course in p

scZoUnD [109]

Answer:

The probability that they will teach different courses is $\frac{2}{3}$ .

Step-by-step explanation:

Sample space is a set of all possible outcomes of an experiment.

In this case we will write the sample space in the form (x, y).

Here <em>x</em> represents the course taught by the first part-time instructor and <em>y</em> represents the course taught by the second part-time instructor.

Denote every course by their first letter.

The sample space is as follows:

S = {(P, P), (P, I), (P, S), (I, P), (I, I), (I, S), (S, P), (S, I) and (S, S)}

The outcomes where the the instructors will teach different courses are:

s = {(P, I), (P, S), (I, P),(I, S), (S, P) and (S, I)}

The probability of an events <em>E</em> is the ratio of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

Compute the probability that they will teach different courses as follows:

$P(\text{Different courses})=\frac{n(s)}{n(S)}=\frac{6}{9}=\frac{2}{3}$

Thus, the probability that they will teach different courses is $\frac{2}{3}$ .

3 0

2 years ago