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wel
4 years ago
13

Tom and Martha are making punch for a party. The proportions given in the recipe are 2 parts of lemonade to 1 part each of pinea

pple juice and ginger ale. If they need 6 liters of punch, how many liters of each of the ingredients should they buy?
Mathematics
1 answer:
OLEGan [10]4 years ago
7 0

Answer:

They need to buy 3 litters of lemonade, 1.5 litters of pineaple juice and 1.5 litters of ginger ale.

Step-by-step explanation:

In order to calculate the amount of each ingredient they need for 6 liters of punch we can first create fractions for each ingredient based on the proportions we were given. This is shown bellow:

2 litters lemonade + 1 litter pineapple juice + 1 litter ginger ale = 4 litters punch

Then we have:

lemonade = 2/4

pineaple juice = 1/4

ginger ale = 1/4

If we want to make 6 liters of punch we can just apply this fractions to know how much of each we need:

lemonade = 6*(2/4) = 12/4 = 3 litters

pineaple juice = 6*(1/4) = 1.5 litters

ginger ale = 6*(1/4)  = 1.5 litters

They need to buy 3 litters of lemonade, 1.5 litters of pineaple juice and 1.5 litters of ginger ale.

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elixir [45]
No. 0.70 is the same as 0.7, for the place value of 7 is the same for both (tenths place). In 0.70, the place value of 0 is the hundredth, which is not needed.


0.70 = 0.7

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3 years ago
What is the distance between (4, 6) and (5, 0)
tresset_1 [31]

Answer:

6 units

Step-by-step explanation:

Given data

x1=4

y1=6

x2=5

y2=0

substitute in the formula for the distance between two points

d=√((x_2-x_1)²+(y_2-y_1)²)

d=√((5-4)²+(0-6)²)

d=√((1)²+(6)²)

d=√36)

d=6

Hence the distance is 6 units

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3 years ago
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3 years ago
Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o
Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0
3 years ago
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