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3 years ago

Nate is 10/9/12 years old. How old will he be in 2/1/12 more years

Mathematics

1 answer:

EastWind [94]3 years ago

5 0

The answer is B. 10 9/12 + 2 1/12= 12 10/12. 12÷2=6 10÷2=5 making the answer 12 5/6

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PLEASE HELP WITH THIS MATH QUESTION

Vanyuwa [196]

Table C is a function because there is one x-value for every y-value.

6 0

3 years ago

Find the angle that the line through the given pair of points makes with the positive direction of the x-axis

alexdok [17]

Answer:

Therefore the angle that the line through the given pair of points makes with the positive direction of the x-axis is 45°.

Step-by-step explanation:

Given:

Let

A(x₁ , y₁) = (1 , 4) and

B( x₂ , y₂ ) = (-1 , 2)

To Find:

θ = ?

Solution:

Slope of a line when two points are given is given bt

$Slope(AB)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Substituting the values we get

$Slope(AB)=\dfrac{2-4}{-1-1}=\dfrac{-2}{-2}=1\\\\Slope=1$

Also Slope of line when angle ' θ ' is given as

$Slope=\tan \theta$

Substituting Slope = 1 we get

$1=\tan \theta$

$\tan \theta=1\\\theta=\tan^{-1}(1)$

We Know That for angle 45°,

tan 45 = 1

Therefore

$\theta=45\°$

Therefore the angle that the line through the given pair of points makes with the positive direction of the x-axis is 45°.

5 0

3 years ago

Will give brainliest.

Dennis_Churaev [7]

Pretty sure it's A. (X times .5 equals Y)

8 0

3 years ago

Write the percent as a fraction or mixed number in simplest form. <br> 62%

suter [353]

= 62%

= 0.62

= 62/100

= 31/50

6 0

3 years ago

Read 2 more answers

A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a

natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student

The sample means = x₁ + x₂ + . . . , x₂₃ / 23

$= \frac{x_1 +x_2 +...x_2_3}{23}$

= 2.3hr

⇒ $\sum xi =2.3 \times 23 = 52.9hrs$

let x₁ , x₂, . . . , x₂₃ arranged in ascending order

Then x₂₃ was 10 and has been changed to 14

i.e x₂₃ increase to 4

Sample mean $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)² ∑ (xi - 2.47)²

Therefore , the standard deviation increases

6 0

3 years ago