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aleksley [76]
3 years ago
13

Please Help me!!! with this problem

Mathematics
1 answer:
dimaraw [331]3 years ago
8 0

9514 1404 393

Answer:

  D  120

Step-by-step explanation:

We are given the relation ...

  48 = 0.40 × sales

Dividing by 0.40 gives the sales for the day.

  sales = 48/0.40 = 120

120 milkshakes were sold that day.

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18h 19min 26s<br> + 25h 45min 48s
snow_tiger [21]

Answer:

2 days 5 minuts 14 seconds

Step-by-step explanation:

 1 1   1 1    1

 18   19   26

+25  45  48

1 24   05  14

4 0
3 years ago
What is the answer to 5/8 1/4?
Aleksandr [31]
5 × 1. = 5
_ × _. = _
8 × 4. = 32


5/32
4 0
3 years ago
According to the bar graph above, which salary range is the least frequent?
Elis [28]

9514 1404 393

Answer:

  D.  $101,000 – $120,000

Step-by-step explanation:

The bar graph is not completely labeled, but in the context of the question it seems safe to assume that the vertical scale can be considered to represent relative frequency.

So, the shortest bar is the one with the lowest frequency. The horizontal scale identifies that as 101-120. If we assume that is salary in thousands of dollars, then Choice D is appropriate.

5 0
3 years ago
Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th
grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

S = \{1,1,2,2,2,3\}

And the probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{6}

P(1) = \frac{1}{3}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{6}

P(2) = \frac{1}{2}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{6}

P(Odd Number) is then calculated as:

P(Odd\ Number) =  P(1) + P(3)

P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}

Take LCM

P(Odd\ Number) = \frac{2+1}{6}

P(Odd\ Number) = \frac{3}{6}

P(Odd\ Number) =  \frac{1}{2}

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

S = \{1,1,2,2,2,3,3\}

The probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{7}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{7}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{7}

P(Odd\ Number) = P(1) + P(3)

P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}

Take LCM

P(Odd\ Number) = \frac{2+1}{7}

P(Odd\ Number) = \frac{3}{7}

Comparing P(Odd Number) before and after

P(Odd\ Number) =  \frac{1}{2} --- Before

P(Odd\ Number) = \frac{3}{7} --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0
3 years ago
Solve for x. Enter the solutions from least to greatest.
Len [333]

Answer:

This is the answer of your question ☺☺

Step-by-step explanation:

(x+7)2 -49=0

2x+14-49=0

2x-35=0

2x=35

x=17.5

7 0
2 years ago
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