All categories

3 years ago

I need help please with this math problem

Mathematics

1 answer:

Arada [10]3 years ago

8 0

When the dilation is centered on the origin, you can multiply each of the individual coordinates by the scale factor.

All the coordinates are ±5 and the scale factor is 1/5. Multiplying those gives ±1. The only selection with all coordinates being ±1 is
.. Selection A.

You can check to see if the signs agree in detail. (They do.)

You might be interested in

How to write 9,009,009 in word form

DaniilM [7]

Nine million nine thousand and nine

7 0

3 years ago

Read 2 more answers

IN NEED OF MORE HELP AND THANK YOU IF YOU ANSWER

Lorico [155]

Answer:

512

Step-by-step explanation:

5.12 converted to a fraction is 512/100

3 0

2 years ago

A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materi

valentina_108 [34]

Answer:

N(x) = 40 - 2x

P(x) = -2x² + 52 x - 240

maximum profit = 13

Step-by-step explanation:

given data

feeder cost = $6

average sell = 20 per week

price = $10 each

solution

we consider here price per feeder = x

and profit per feeder id here formula = x - 6

so that here

total profit will be

P (x) = ( x - 6 ) Nx

here N(x) is number of feeders sold at price = x

so formula for N (x) is here

N(x) = 20 - 2 ( x - 10 )

N(x) = 40 - 2x

so that

P(x) = (x-6) ( 40 - 2x)

P(x) = -2x² + 52 x - 240

since here

a = -2

b = 52

c = -240

a < 0

so quadratic function have maximum value of c - $\frac{b^2}{4a}$

so it will be

maximum value = -240 - $\frac{52^2}{4(-2)}$

maximum value = 98

so here maximum profit attained at

x = $\frac{-b}{2a}$

x = $\frac{-52}{2(-2)}$

x = 13

maximum profit = 13

3 0

3 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

3 years ago

Which graph shows the solution set for 2 x + 3 greater-than negative 9?

sp2606 [1]

I''m pretty sure it's A

6 0

3 years ago