All categories

4 years ago

To keep the image in proportion what do you have to do?

Mathematics

1 answer:

hoa [83]4 years ago

5 0

Answer:

Press shift while you drag a corner handle to prevent stretching and keep your picture in proportion. Don't drag from any of the side handles; this will distort the picture even if you do press Shift!

Step-by-step explanation:

You might be interested in

Round the fraction 7 4/5 to the nearest whole number

Over [174]

Im not sure if you are saying 74/5 or 7 and 4 fifths but in the case you'r saying 74/5 you would first round to see the highest number that 5 can go to without being a decimal. Which would be 70 so you would count how many times 5 can go into 70, which would be 14, so the left over is 4/5 so you would check to see if it can be simplified which it can't so the final answer is 14 and 4/5

3 0

3 years ago

Read 2 more answers

Drey makes $10 an hour plus $15 an hour for every hour of overtime. Overtime hours are any hours more than 40 hours for the week

alekssr [168]

Drey makes $10 an hour plus $15 an hour for every hour of overtime. Overtime hours are any hours more than 40 hours for the week.

Part A: Create an equation that shows the amount of wages earned, W, for working x hours in a week when there is no overtime.

If there is no overtime, the equation is W = 10x.

Part B: Create an equation that shows the amount of wages earned, S, for working y hours of overtime. Hint: Remember to include in the equation the amount earned from working 40 hours.

At $10 per hour for 40 hours, the total is $400.

S = 15y + 400

Part C: Drey earned $490 in 1 week. How many hours (regular plus overtime) did he work?

490 = 15y + 400

490 - 400 = 15y

90 = 15y

90/15 = y

6 = y

Drey worked 46 hours in all.

4 0

3 years ago

Can somebody help me please

SpyIntel [72]

<h2>Answer:</h2><h3>f(4) = 122</h3>

________________________________________________

<h3>Calculate</h3>

$Substitute\ x=4\ into\\ f(x)=7x^2+4x-6$

_________________________________________________

<h3>Substitute</h3>

$f(4)=7\times 4^2+4\times 4-6$

___________________________________________________

<h3>Calculate the power</h3>

$f(4)=7\times 16+4\times 4-6$

___________________________________________________

<h3>Calculate the product or quotient</h3>

$f(4)=112+4\times 4-6\\ f(4)=112+16-6$

___________________________________________________

<h3>Calculate the sum or difference</h3>

$f(4)=128-6\\ f(4)=122$

<em>I hope this helps you</em>

6 0

3 years ago

Suppose n people, n ≥ 3, play "odd person out" to decide who will buy the next round of refreshments. The n people each flip a f

blondinia [14]

Answer:

Assume that all the coins involved here are fair coins.

a) Probability of finding the "odd" person in one round: $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}$ .

b) Probability of finding the "odd" person in the $k$ th round: $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left( 1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}$ .

c) Expected number of rounds: $\displaystyle \frac{2^{n - 1}}{n}$ .

Step-by-step explanation:

To decide the "odd" person, either of the following must happen:

There are $(n - 1)$ heads and $1$ tail, or
There are $1$ head and $(n - 1)$ tails.

Assume that the coins here all are all fair. In other words, each has a $50\,\%$ chance of landing on the head and a

The binomial distribution can model the outcome of $n$ coin-tosses. The chance of getting $x$ heads out of

The chance of getting $(n - 1)$ heads (and consequently, $1$ tail) would be $\displaystyle {n \choose n - 1}\cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left(\frac{1}{2}\right)^{n - (n - 1)} = n\cdot \left(\frac{1}{2}\right)^n$ .
The chance of getting $1$ heads (and consequently, $(n - 1)$ tails) would be $\displaystyle {n \choose 1}\cdot \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{n - 1} = n\cdot \left(\frac{1}{2}\right)^n$ .

These two events are mutually-exclusive. $\displaystyle n\cdot \left(\frac{1}{2}\right)^n + n\cdot \left(\frac{1}{2}\right)^n = 2\,n \cdot \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ would be the chance that either of them will occur. That's the same as the chance of determining the "odd" person in one round.

Since the coins here are all fair, the chance of determining the "odd" person would be $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}$ in all rounds.

When the chance $p$ of getting a success in each round is the same, the geometric distribution would give the probability of getting the first success (that is, to find the "odd" person) in the $k$ th round: $(1 - p)^{k - 1} \cdot p$ . That's the same as the probability of getting one success after $(k - 1)$ unsuccessful attempts.

In this case, $\displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ . Therefore, the probability of succeeding on round $k$ round would be

$\displaystyle \underbrace{\left(1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}}_{(1 - p)^{k - 1}} \cdot \underbrace{n \cdot \left(\frac{1}{2}\right)^{n - 1}}_{p}$ .

Let $p$ is the chance of success on each round in a geometric distribution. The expected value of that distribution would be $\displaystyle \frac{1}{p}$ .

In this case, since $\displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ , the expected value would be $\displaystyle \frac{1}{p} = \frac{1}{\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}}= \frac{2^{n - 1}}{n}$ .

7 0

3 years ago

A gambler claims that horses in a horse race that start closer to the rail have an advantage over horses that start further from

k0ka [10]

Answer:

Hence, the p-value ( 0.699986) > 0.05, FAIL TO REJECT THE NULL HYPOTHESIS

there is no significant evidence that starting position is a factor in horse races

Step-by-step explanation:

Given the data in the question;

Starting position: 1 2 3 4 5 6

# of Winners: 25 22 18 19 21 15

Mean x" (E) = (25+22+18+19+21+15)/6 = 120/6 = 20

Observed/Expected value table;

O E ((O-E)²) / E

25 20 1.25

22 20 0.2

18 20 0.2

19 20 0.05

21 20 0.05

15 20 1.25

∑ 3

so using chi square = ∑(((O-E)²) / E) = 3

DF = n-1 = 6-1 = 5

then, the critical chi square value is;

significance level = 0.05

chi square( critical ) = 11.07049769

from table;

p - value = 0.699986

Hence, the p-value ( 0.699986) > 0.05, FAIL TO REJECT THE NULL HYPOTHESIS

there is no significant evidence that starting position is a factor in horse races

8 0

3 years ago