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2 years ago

7

An elevator ascends, or goes up, at a rate of 750 feet per minute. Is the height to which the elevator ascends proportional to t

he number of minutes it takes to get there? (Examples 1-3)

1 answer:

Marina CMI [18]2 years ago

5 0

Answer:

1.)750 2.) 1,500 3.)2,250 4.) 3,000

Step-by-step explanation:

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Number 8 please help look at the picture

kkurt [141]

You have to convert kilograms to grams or grams to kilograms. Conversion is; Kilo Hecto Deca (base-which in this case is grams) deci centi milli.

5 0

2 years ago

what are the minimum first quartile median third quartile and the maximum of the data set? 9,20,4,18,4,18,20,9

den301095 [7]

<u>Answers</u>
1. Minimum = 4
2. First quartile = 6.5
3. Median = 13.5
4. Third quartile = 19
5. Maximum = 20

<u>Explanation</u>
To calculate the measure of central tendency, you first arrange the set of the data in ascending order.
The set of data given will be;
4, 4, 9, 9, 18, 18, 20, 20.

Part 1:
The minimum value of the data is 4.

Part 2:
The first quatile is the median of the lower half which is comprised by:
4, 4, 9, 9

1st quartile = (4+9)÷2
= 13÷2
= 6.5

Part 3:
Median of the data is;

Median = (9+18)÷2
=27÷2
= 13.5

Part 4:
3rd quartile is the median of the upper half which comprises of;
18, 18, 20, 20.

3rd quartile = (18+20)÷2
= 48÷2
= 19

Part 5
The maximum of the set of data is 20.

7 0

2 years ago

A grocery store sells a bag of 5 lemons for 2$. watnis the unit cost of each lemon in the bag

Ostrovityanka [42]

Answer:

$0.4

Step-by-step explanation:

Unit cost = cost / number of items

Unit cost = 2/5

2 divided by 5

3 0

2 years ago

There are (49)5 • 40 books at the library. What is the total number of books at the library? (4 points)

madreJ [45]

5 times 40 times 49 equals 9800 because you do 5 time 4 and that's 20 and you add a 0 and you multiply by 49 to get 9800.

5 0

2 years ago

Read 2 more answers

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

2 years ago