I need to do 41,43,45,and 47

If 3 ice cream cones cost 8.25 how much do 2 ice cream cones cost

Fed [463]

If 3 ice cream cones cost 8.25 how much do 2 ice cream cones cost?

1 cone cost

8.25 : 3 = 2.75

2 cones cost

2.75 * 2 = 5.5

or

(8.25 : 3) * 2 = 5.5

4 0

3 years ago

Read 2 more answers

The manager of a supermarket wants to obtain information about the proportion of customers who dislike a new policy on cashing c

alex41 [277]

Answer:

n = 61 costumers

Step-by-step explanation:

For calculating the number of costumers he should sample we use the next equation:

$n = \frac{z_{1-\alpha/2}^{2}p(1-p)}{E^{2} }$

Where E is the error that we are prepared to accept, in this case E = 0.15

How we don't know the value of p, we can estimate it like p = 0.5

∝ = 1-0.98 = 0.02

1-∝/2 = 0.99

$z_{0.99} = 2.33$

$n = \frac{(2.33^{2})(0.5)(1-0.5)}{0.15^{2} }$

n = 60.32 costumers

n ≈ 61 costumers

4 0

3 years ago

Find the interquartile range of the data.

likoan [24]

The answer is 14. Here is how I work it out.

First we are going have to identify quartile 1 and quartile 3.

So after putting the numbers in order from least to greatest mark the number with a half way point.

This is optional but it will help us spot the sections better.

31,33,35,41,43,|46,48,49,49,50

Next put parentheses around the remaining groups of numbers.

(31,33,35,41,43,)|(46,48,49,49,50)

For the next step we have to find the median of each group.

(31,33,35,41,43,)|(46,48,49,49,50)

The median of each group, are called the quartiles, the median of the lower half is quartile 1, and the median of the upper half is quartile 3.

Now subtract quartile 1 from quartile 3.

49 – 35 = 14

So the interquartile range of this data set is 14.

4 0

3 years ago

Lowest common multiple of 9

ch4aika [34]

Is 3.

3×3=9

Hope this helps!

3 0

3 years ago

According to DeMorgan 's theorem, the complement of W · X + Y · Z is W' + X' · Y' + Z'. Yet both functions are 1 for WXYZ = 1110

sergeinik [125]

Answer:

The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.

Step-by-step explanation:

According to DeMorgan's Theorem:

(W.X + Y.Z)'

(W.X)' . (Y.Z)'

(W'+X') . (Y' + Z')

Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.

For the original function:

(W . X + Y . Z)'

= (1 . 1 + 1 . 0)

= (1 + 0) = 1

For the compliment:

(W' + X') . (Y' + Z')

=(1' + 1') . (1' + 0')

=(0 + 0) . (0 + 1)

=0 . 1 = 0

Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.

Without the parenthesis the compliment equation looks like this:

W' + X' . Y' + Z'

1' + 1' . 1' + 0'

0 + 0 . 0 + 1

Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.

Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.

3 0

3 years ago