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e-lub [12.9K]
3 years ago
5

megan is going to make a drink using the instrutions below mix 2 parts of fruit juice with 5 parts if sparkling water. megan has

180ml of fruit juice and 400ml of sparkling water. what is the greatest amount of the drink can megan can make?
Mathematics
1 answer:
OLEGan [10]3 years ago
5 0

Answer:

160 mL of fruit juice and 400 mL of sparkling water

Step-by-step explanation:

According to the statement we have that the ratio between fruit juice and sparkling water is 2/5.

Therefore, we could also say that:

200/500

with it could be done with 200 mL of fruit juice and 500 mL of sparkling water, so for 400 mL, how much would it be:

400 * 200/500 = 160

Therefore, the largest amount of the drink can be made with 160 mL of fruit juice and 400 mL of sparkling water.

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Expand: -7/3(x-1)+2: 13/3-7x/3
expand: 4(2-x): 8-4x
13/3 - 7x/3 =8-4x
13/3 -7x/3 -13/3 = 8-4x -13/3
-7x/3=-4x+11/3
-7x/3 * 3= -4x*3+11/3*3
-7x=-12x+11
-7x+12x=-12x+11+12x
5x=11
5x/5=11/5
x=11/5

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Help me find the slope! (brainliest) <br><br> Jorge lost 18 pounds in 3 months.
krek1111 [17]

Do you have a graph for your question?

4 0
3 years ago
Read 2 more answers
Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> In x = 3
mixer [17]

Answer:

The solution is:

x = e^{3} = 20.09

Step-by-step explanation:

The first step to solve this equation is placing everything with the logarithmicto one side of the equality, and everything without the exponential to the other side. So

\ln{x} = 3

It already is in the desired format.

Now, we have that, since e and ln are inverse operations

e^{\ln{a}} = a

So, we apply the exponential to both sides of the equality

e^{\ln{x}} = e^{3}

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7 0
3 years ago
A book claims that more hockey players are born in January through March than in October through December. The following data sh
astra-53 [7]

Answer:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the​ year

H1: There is a difference between birthdates distributed throughout the​ year

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total}{4}

And replacing we got:

E_{1} =\frac{67+56+30+37}{4}=47.5

And now we can calculate the statistic:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

3 0
4 years ago
Please help me please.
ValentinkaMS [17]
I think it's either 8 (16/2=8) OR 32 (16x2=32)
7 0
3 years ago
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