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Alex787 [66]
4 years ago
6

Ralf and Susie share $57 in the ratio 2:1. (a) Calculate the amount Ralf receives.

Mathematics
2 answers:
Mnenie [13.5K]4 years ago
4 0

Answer:

<h3>.Total ratio 2+1 = 3</h3>

<h3><u>Ralf</u><u> </u></h3>

<h3>2/3 × 57</h3><h3>Divide 57 by 3</h3><h3>( 2 × 19)</h3><h2><u>=</u><u> </u><u>3</u><u>8</u><u>.</u><u>.</u><u>.</u><u>.</u></h2><h3><u>Ralf</u><u> </u><u>recieves</u><u> </u><u>$</u><u> </u><u>3</u><u>8</u></h3>
Semmy [17]4 years ago
4 0

Answer:

Raff = 38

Suzie = 19

Step-by-step explanation:

Raff gets 2x

Suzie gets x

2x + x = 57 dollars

3x = 57 dollars

x = 19 dollars

2x = 38 dollars

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Which number line correctly shows the location or approximate location of the three numbers?
GREYUIT [131]
<h2>Hello!</h2>

The answer is: D. is the fourth picture.

<h2>Why?</h2>

First, we need to make these numbers entire or decimal numbers, in order to see where are they located:

2\sqrt{5} =4.47\\\frac{6}{7}*\pi =2.69\\\sqrt{16} =4

So, we should look for the image that shows the following order it: 2.69, 4 and 4.47

or

\frac{6}{7} *\pi

\sqrt{16} =4

2\sqrt{5}

Then, we just need to find the image where they are properly located/ordered:

So, the correct image is D. is the fourth picture.

Have a nice day!

3 0
4 years ago
Raul calculated that he would spend $125 on school supplies this year. He actually spent $87.50 on school supplies. What is Raul
Eddi Din [679]
<span>87.5 is 70% of 125 so the percent of error is 30%</span>
3 0
3 years ago
Read 2 more answers
How do I expand 3x(2-10x)
Hunter-Best [27]

Answer:

(3x + 5)^2 expands into two binomials: (3x + 5)(3x +5)


from there, you foil them:

(3x + 5)(3x + 5)

multiply the first terms ... 9x^2

the inside terms ... 9x^2 + 15x

outside terms ... 9x^2 + 15x + 15x

last terms ... 9x^2 + 15x + 15x + 25

then add like terms ... 9x^2 + 30x + 25


3 0
3 years ago
Read 2 more answers
i-Ready Which kind of triangle is shown? equilateral triangle acute triangle right triangle obtuse triangle​
AveGali [126]

Answer:

right triangle

Step-by-step explanation:

Notice it has a right angle (90º) in its corner. Therefore, it is a right triangle.

Right triangles look like this: ⊿

7 0
3 years ago
A simple random sample of size nequals=8181 is obtained from a population with mu equals 77μ=77 and sigma equals 27σ=27. ​(a) De
ivanzaharov [21]

Answer:

a) P(\bar X>81.5)=1-0.933=0.067

b) P(\bar X

c) P(73.4  

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

X \sim N(\mu=77,\sigma=27)  

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(77,\frac{27}{\sqrt{81}})

Part a

We want this probability:

P(\bar X>81.5)=1-P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X >81.5)=1-P(Z

P(\bar X>81.5)=1-0.933=0.067

Part b

We want this probability:

P(\bar X\leq 69.5)

If we apply the formula for the z score to our probability we got this:

P(\bar X \leq 69.5)=P(Z\leq \frac{69.5-77}{\frac{27}{\sqrt{81}}})=P(Z

P(\bar X\leq 69.5)=0.0062

Part c

We are interested on this probability

P(73.4  

If we apply the Z score formula to our probability we got this:

P(73.4

=P(\frac{73.4-77}{\frac{27}{\sqrt{81}}}

And we can find this probability on this way:

P(-1.2

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-1.2

3 0
4 years ago
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