Negative magnitude rotation is

Which equation describes the pattern of the data in the table?

slega [8]

Answer:

Option A. y = 4x

Step-by-step explanation:

From the question given above, the following data were obtained:

X >> Y

3 >> 12

4 >> 16

5 >> 20

6 >> 24

7 >> 28

Next, we shall use each of the options to obtain the first two value of y to which will correspond to the table above. This is illustrated below:

For Option A:

1. y = 4x

x = 3

y = 4 × 3

y = 12

2. y = 4x

x = 4

y = 4 × 4

y = 16

For Option B:

1. y = 4x + 12

x = 3

y = 4(3) + 12

y = 12 + 12

y = 24

2. y = 4x + 12

x = 4

y = 4(4) + 12

y = 16 + 12

y = 28

For Option C:

1. y = ¼ x

x = 3

y = ¼ × 3

y = ¾

2. y = ¼ x

x = 4

y = ¼ × 4

y = 1

For Option D:

1. y = ¼x + 12

x = 3

y = ¼(3) + 12

y = ¾ + 12

y = 51/4

2. y = ¼x + 12

x = 4

y = ¼(4) + 12

y = 1 + 12

y = 13

From the calculations made above, only option A ie. y = 4x correspond to the data given in the table above.

5 0

3 years ago

Need help................

Triss [41]

827537517172763 kahvxjx

6 0

3 years ago

Starting at 6 a.m. every morning, Matilda receives text messages on her cell phone from her mother, her best friend, and her bro

Dafna1 [17]

Answer:

a) 0.0013 = 0.13% probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

b) $1.6 \times 10^{-9}$ probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

Step-by-step explanation:

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A) Find the probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

Two from the best friend:

Her best friend sends a message once every 10 minutes.

From 6 to 7:30, there is an hour and a half, that is, 90 minutes, so the mean for her best friend is $\mu = \frac{90}{10} = 9$

Two messages is P(X = 2). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-9}*9^{2}}{(2)!} = 0.0050$

Two from the mother:

Message every hour = 60 minutes. So $\mu = \frac{90}{60} = 1.5$ . This is P(X = 2).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-1.5}*1.5^{2}}{(2)!} = 0.2510$

Two of her best friend and two of her mother:

Independent events, so the probability of both happening is the multiplication of their separate probabilities.

$p = 0.005*0.251 = 0.0013$

0.0013 = 0.13% probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

B) With a chance of 75% a text message contains a typo independent of the sender. Find the probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

In 3 hours, she is expected to receive:

3*60/10 = 18 messages from her best friend.

3*60/60 = 3 messages from her mother.

3*60/30 = 6 messages from her brother.

In total, 27 messages.

75% probability of a typo, so $\mu = 0.75*27 = 20.25$

This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-20.25}*20.25^{0}}{(0)!} = 1.6 \times 10^{-9}$

$1.6 \times 10^{-9}$ probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

8 0

3 years ago

Simplify the expression:<br> 7r–3r+9r

Arte-miy333 [17]

Answer:

13r

Step-by-step explanation:

all like variables so add the coefficient

7+(-3)+9

7-3+9

13

=13r

8 0

3 years ago

PLEASE HELP ME WITH THIS PROBLEM, SYSTEM OF LINEAR EQUATIONS, 30 POINTS!!!!

Alex Ar [27]

Answer:

You can either use substation or you can use elimination on both of the problems

Step-by-step explanation:

5 0

3 years ago