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Katyanochek1 [597]
2 years ago
15

The cost of one apple and one banana at the school canteen is $1 and the cost of 3 apples and 2 bananas is $2.40. Find the cost

of a single banana, using simultaneous equation, elimination method. Show full working out.
Mathematics
2 answers:
Anastaziya [24]2 years ago
8 0

Answer:

I don't know

I don't know

marta [7]2 years ago
7 0
Let the apple equal to x and let the banana equal to y

Equation one is x+y= 1

Equation two is 3x+2y= 2.40

Because we are doing the elimination method, multiply all of equation one by 3


3x+3y=3
3x+2y=2.40
(subtract the equations from one another)


So now you have y=60 cents

So, one banana costs 60 cents
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Answer:

the  amount that pay to the agency is $1,091.67

Step-by-step explanation:

The computation of the amount that pay to the agency is as follows:

= Amount paid per year ÷ number of months × 25%

= $52,400 ÷ 12 months × 25%

= $1,091.67

Hence, the  amount that pay to the agency is $1,091.67

We simply applied the above formula so that the correct amount could come

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Admission for children is $7 and $10 for adults the event was attended by 700 people and total amount collected was $6400 there
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There were 500 adults and 200 children.
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The Salk polio vaccine experiment in 1954 focused on the effectiveness of the vaccine in combating paralytic polio. Because it w
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Answer:

Step-by-step explanation:

Hello!

The variables of interest are:

X₁: Number of cases of polio observed in kids that received the placebo vaccine.

n₁= 201299 total children studied

x₁= 110 cases observed

X₂: Number of cases of polio observed in kids that received the experimental vaccine.

n₂= 200745 total children studied

x₂= 33 cases observed

These two variables have a binomial distribution. The parameters of interest, the ones to compare, are the population proportions: p₁ vs p₂

You have to test if the population proportions of children who contracted polio in both groups are different: p₂ ≠ p₁

a)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.05

Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }

Sample proportion placebo p'₁= x₁/n₁= 110/201299= 0.0005

Sample proportion vaccine p'₂= x₂/n₂= 33/200745= 0.0002

Pooled sample proportion p'= (x₁+x₂)/(n₁+n₂)= (110+33)/(201299+200745)= 0.0004

Z_{H_0}= \frac{(0.0002-0.0005)-0}{\sqrt{0.0004[\frac{1}{201299} +\frac{1}{200745} ]} }= -4.76

This test is two-tailed, using the critical value approach, you have to determine two critical values:

Z_{\alpha/2}= Z_{0.025}= -1.96

Z_{1-\alpha /2}= Z_{0.975}= 1.96

Then if Z_{H_0} ≤ -1.96 or if Z_{H_0} ≥ 1.96, the decision is to reject the null hypothesis.

If -1.96 < Z_{H_0} < 1.96, the decision is to not reject the null hypothesis.

⇒ Z_{H_0}= -4.76, the decision is to reject the null hypothesis.

b)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.01

Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }

The value of Z_{H_0}= -4.76 doesn't change, since we are working with the same samples.

The only thing that changes alongside with the level of significance is the rejection region:

Z_{\alpha /2}= Z_{0.005}= -2.576

Z_{1-\alpha /2}= Z_{0.995}= 2.576

Then if Z_{H_0} ≤ -2.576or if Z_{H_0} ≥ 2.576, the decision is to reject the null hypothesis.

If -2.576< Z_{H_0} < 2.576, the decision is to not reject the null hypothesis.

⇒ Z_{H_0}= -4.76, the decision is to reject the null hypothesis.

c)

Remember the level of significance (probability of committing type I error) is the probability of rejecting a true null hypothesis. This means that the smaller this value is, the fewer chances you have of discarding the true null hypothesis. But as you know, you cannot just reduce this value to zero because, the smaller α is, the bigger β (probability of committing type II error) becomes.

Rejecting the null hypothesis using different values of α means that there is a high chance that you reached a correct decision (rejecting a false null hypothesis)

I hope this helps!

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