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3 years ago

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Noah spent some of his allowance money on Monday. He spent 1/6 of his allowance on Tuesday and 3/6 of it on Friday. Noah has non

e of his allowance money left. What fraction of his allowance did he spend on Monday.

1 answer:

Harrizon [31]3 years ago

5 0

Answer:

2/6 (also simplified to 1/3)

Step-by-step explanation:

1 + 3 = 4

6 - 4 = 2

divide 2 by 6 and you get 2/6

hopefully this helps ;)

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This Venn diagram shows sports played by 10 students.

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B $\frac{7}{10}$

Step-by-step explanation:

Event A = 2

Event B = 3

Ian and Ella play both so that would add 2 more people.

Total who play soccer or basketball = 7

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$= \frac{7}{10}$

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Nathan is going to invest in an account paying an interest rate of 4.8% compounded continuously. How much would Nathan need to i

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How do you do this question?

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Answer:

C. y₂ = (1 + (t/n))²

Step-by-step explanation:

yₙ₊₁ = yₙ + Δt F(tₙ, yₙ)

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yₙ₊₁ = yₙ + (t/n) yₙ

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y₂ = 1 + (t/n) + (t/n) (1 + (t/n))

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y₂ = 1 + 2(t/n) + (t/n)²

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3 years ago

Solve the inequality for V.<br><br> -8.3 _>_ V - 1.9

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First keep in mind that the given value is negative and that it is greater than or equal to whatever '<em>v</em>' is.

$-8.3 \geq v - 1.9$

When solving for a variable in any equation, you do something to both sides in order to keep it equal. Here, <em>v</em> is being subtracted by 1.9; therefore we can add 1.9 to both sides in order to isolate <em />the variable.

$-8.3 + 1.9 \geq v - 1.9 + 1.9$
$-6.4 \geq v$

Despite not needing a value for the question, it is worth noting that since this is an inequality, <em>v </em>can be any value from -6.4 to ∞ in order to make it true.

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2 years ago

A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po

igor_vitrenko [27]

Answer:

a) For this case the value of the significanceis $\alpha=0.05$ and $\alpha/2 =0.025$ , we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

$z_{\alpha/2} =1.96$

If the calculated statistic $|z_{calc}| >1.96$ we can reject the null hypothesis at 5% of significance

b) Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8$

c) $z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708$

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation

$X_{1}=170$ represent the number of people with the characteristic 1

$X_{2}=110$ represent the number of people with the characteristic 2

$n_{1}=200$ sample 1 selected

$n_{2}=150$ sample 2 selected

$p_{1}=\frac{170}{200}=0.85$ represent the proportion estimated for the sample 1

$p_{2}=\frac{110}{150}=0.733$ represent the proportion estimated for the sample 2

$\hat p$ represent the pooled estimate of p

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

$\alpha=0.05$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

a.State the decision rule.

For this case the value of the significanceis $\alpha=0.05$ and $\alpha/2 =0.025$ , we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

$z_{\alpha/2} =1.96$

If the calculated statistic $|z_{calc}| >1.96$ we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8$

c. Compute the value of the test statistic.

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708$

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

5 0

3 years ago