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3 years ago

Is how many sixth graders attend your school a statistical question?

Mathematics

2 answers:

HACTEHA [7]3 years ago

8 0

Yes, any question that involves numbers would be a statistical question.

Bogdan [553]3 years ago

5 0

It depends, if you're asking a multitude of Junior High's it could be, here's an example of a question that would be statistical though.

<span>How many minutes do 6th grade students typically spend on homework each week?</span>

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I'm supposed to simplify this.

tekilochka [14]

Answer:

Step-by-step explanation:

the proof of the answer is seen in the diagram

4 0

3 years ago

Which of the following expressions shows how to rewrite 5 − 6 using the additive inverse and displays the expression correctly o

Andru [333]

The correct display of the expression on a number line is number line (c)

<h3>How to determine the expression that shows how to rewrite 5 − 6 using the additive inverse?</h3>

The complete question is added as an attachment

The expression is given as:

5 - 6

Put 6 in a bracket.

So, we have:

5 - 6 = 5 - (6)

Express 6 as --6

So, we have:

5 - 6 = 5 - -(-6)

Rewrite as:

5 - 6 = 5 + (-6)

The above means that the arrow on the number line must start from 5 and moves in the negative direction in 6 units

The correct display of the expression on a number line is number line (c)

Read more about number lines at:

brainly.com/question/24644930

#SPJ1

7 0

1 year ago

Samira used the calculator to find 254.3 × 2.5. Which statement best describes her work? Samira is correct. Samira is incorrect.

alekssr [168]

Answer:

A samira is correct

Step-by-step explanation:

Because she did her own expression right, now she has to solve

4 0

3 years ago

Read 2 more answers

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$