Using a line of best fit, which value would most likely be predicted for an x-value of 16?

Need help on this one

Artist 52 [7]

Answer:

12

Step-by-step explanation: it’s right

8 0

3 years ago

Help me please asap

LiRa [457]

Answer:

x = 20

Step-by-step explanation:

AE is an angle bisector, which means it cuts that triangle into 2 equal halves. So this means ∠BAE and ∠EAC should be equal to each other.

Set them equal to each other and solve for x.

∠BAE = ∠EAC

x + 30 = 3x - 10

40 = 2x

x = 20

3 0

3 years ago

The scale on a map is 2 centimeters= 50 kilometres.Two rivers on a map are located 9.3 centimeters apart.What is The actual dist

Yuki888 [10]

Answer:

The actual distance between the two rivers is 232.5 kilometers.

Step-by-step explanation:

GIven:

The scale on a map is 2 centimeters= 50 kilometres.Two rivers on a map are located 9.3 centimeters apart.

Now, to find the actual distance between the two rivers.

Let the actual distance between the two rivers is $x.$

The two rivers on the map is located apart of 9.3 centimeters.

According to the scale on the map is 2 centimeters = 50 kilometers.

So, 2 centimeters is equivalent to 50 kilometers.

Thus, 9.3 centimeters is equivalent to $x.$

Now, to solve by using cross multiplication method:

$\frac{2}{50} =\frac{9.3}{x}$

By cross multiplying we get:

$2x=465$

Dividing both sides by 2 we get:

$x=232.5\ kilometers.$

Therefore, the actual distance between the two rivers is 232.5 kilometers.

6 0

3 years ago

Eric teaches ceramics in his studio.he estimates that one out of every five people who call for information about a class will s

Vlada [557]

The probability that four or fewer of the people who called will sign up for a class = 0.9805

For given question,

Eric estimates that one out of every five people who call for information about a class will sign up for the class.

Last week he receive nine calls.

We need to find the probability that four or fewer of the people who called will sign up for a class.

Total number of calls = 9

⇒ n = 9

Since one out of every five people who call for information about a class will sign up for the class.

the probability of success (p) = 1/5

= 0.2

and the probability of failure (q) = 1 - p

= 1 - 0.2

= 0.8

To find the probability that four or fewer of the people who called will sign up for a class.

So, x would take values 0, 1, 2, 3, 4

Using Binomial principal,

For x = 0,

$P(x=0)= ~^9C_0(0.2)^0(0.8)^{9-0}\\\\P(x=0)=0.13422$

For x = 1,

$P(x=1)= ~^9C_1(0.2)^1(0.8)^{9-1}\\\\P(x=1)=0.30199$

For x = 2,

$P(x=2)= ~^9C_2(0.2)^2(0.8)^{9-2}\\\\P(x=2)=0.30199$

For x = 3,

$P(x=3)= ~^9C_3(0.2)^3(0.8)^{9-3}\\\\P(x=3)=0.17616$

For x = 4,

$P(x=4)= ~^9C_4(0.2)^4(0.8)^{9-4}\\\\P(x=4)=0.06606$

So, the required probability would be,

P = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)

P = 0.1342 + 0.3020 + 0.3020 + 0.1762 + 0.0661

P = 0.9805

Therefore, the probability that four or fewer of the people who called will sign up for a class = 0.9805

Learn more about the probability here:

brainly.com/question/3679442

#SPJ4

3 0

1 year ago

Mai biked 5 1/4 miles today, and Noah biked 2 1/2 miles. How many times the length of Noah’s bike ride was Mai’s bike ride?

OlgaM077 [116]

Answer:

The length of Mai's bike ride was 2.1 times the length of Noah's ride.

Step-by-step explanation:

Mai biked 5 1/4 miles today

So he biked, in miles:

$5 + \frac{1}{4} = \frac{5*4 + 1}{4} = \frac{20+1}{4} = \frac{21}{4}$

Noah biked 2 1/2 miles.

So, in miles, he biked:

$2 + \frac{1}{2} = \frac{2*2 + 1}{2} = \frac{4+1}{2} = \frac{5}{2}$

How many times the length of Noah’s bike ride was Mai’s bike ride?

We divide the Mai distance by Noah's distance. In a division of fractions, we multiply the numerator by the inverse of the denominator. So

$\frac{\frac{21}{4}}{\frac{5}{2}} = \frac{21}{4} \times {2}{5} = \frac{21*1}{2*5} = \frac{21}{10} = 2.1$

The length of Mai's bike ride was 2.1 times the length of Noah's ride.

5 0

3 years ago