All categories

3 years ago

What is the scale factor?

Mathematics

2 answers:

erastova [34]3 years ago

6 0

Answer:

Step-by-step explanation:

2 x 4 = 8

4 x 4 = 16

So SF = 4 (enlargement)

KengaRu [80]3 years ago

4 0

Scale factor in this question is 4

You might be interested in

How does the graph of f(x) = (x − 9)4 − 3 compare to the parent function g(x) = x4?

Paraphin [41]

Answer:

The graph is shifted right by 9 units and down by 3 units

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

PLEASE HELP!! i cant figure out how to do it

vodomira [7]

Answer:

168 and 12

Step-by-step explanation:

use fractions for this think of it as 14/15x and 1/15x and x is 180 and solve from there

3 0

2 years ago

Ms. Clint is comparing the sales of sweaters and gloves at her store for the past ten winter weeks. Select the true statement ba

lianna [129]

Answer:

i would say b

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Need help with this asap!

Brums [2.3K]

Answer:

The answer to your question is b = 8.06 cm

Step-by-step explanation:

Data

hypotenuse = c = 9 cm

long leg = b = ?

short leg = a = 4 cm

Process

This is a right triangle so we must use the Pythagorean theorem to find the length of the long leg (b)

c² = a² + b²

-Solve for b²

b² = c² - a²

-Substitution

b² = 9² - 4²

-Simplification

b² = 81 - 16

b² = 65

-Result

b = 8.06 cm

4 0

3 years ago

Read 2 more answers

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago