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3 years ago

A^10 x a^4 help .....

Mathematics

1 answer:

nignag [31]3 years ago

3 0

Answer:

Solution

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Slope of 1,-7 and -3-4

3241004551 [841]

$\large \mathfrak{Solution : }$

Slope of the given line is :

$\dfrac{y_2 - y_1}{x_2 - x_1}$

where ,

$x_2 = 1$

$x_1 = - 3$

$y_2 = - 7$

$y_1 = - 4$

let's solve :

$\dfrac{ - 7 - ( - 4)}{1 - ( - 3)}$

$\dfrac{ - 3}{4}$

Slope = -3 / 4

4 0

3 years ago

Solve for b2 in A=1/2 h(b1+b2), if A=16, h=4, and b1=3

icang [17]

Answer:

$b_{2}$ = 5

Step-by-step explanation:

Substitute the given values into the formula

16 = $\frac{1}{2}$ × 4(3 + $b_{2}$ )

16 = 2(3 + $b_{2}$ ) ( divide both sides by 2 )

8 = 3 + $b_{2}$ ( subtract 3 from both sides )

5 = $b_{2}$

4 0

3 years ago

A random representative survey of 100 seventh-graders in Washington Middle School shows that 20 students read 5 or more books ov

DaniilM [7]

20 out of 100 read 5 or more.

20/100 = 0.2

Now multiply the total students by 0..2:

275 x 0.2 = 55

55 people read 5 or more books.

8 0

3 years ago

An algebraic expression that may contain numbers, variables, and exponents.

erica [24]

Answer:

3x^2 + 7xy + 5

Step-by-step explanation:

3 0

3 years ago

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

4 years ago