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3 years ago

15

A bag holds 1 blue block and 3 yellow blocks. How many green blocks should be added to the bag to make the probability of random

ly selecting a green block from the bag equal to 1/2 ?

2 answers:

Klio2033 [76]3 years ago

8 0

Answer:

Step-by-step explanation:

ITS 4

Drupady [299]3 years ago

6 0

$n$ - the number of green blocks which is equal to the number of favourable events
$1+3+n=4n$ - the number of all blocks, which is equal to the number of all possible outcomes

$P(A)=\dfrac{n}{4+n}=\dfrac{1}{2}\\
1\cdot( 4+n)=2\cdot n\\
4+n=2n\\
2n-n=4\\
n=4
$

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Describe the solution of g(x) shown in the graph.

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Answer:

All real solutions

Step-by-step explanation:

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It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir

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We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

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Now, we find the margin of error M as such

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In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

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Media experts claim that daily print newspapers are declining because of Internet access. Listed below, from left to right and

yKpoI14uk [10]

Answer:

(a) The median is 1478

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The test statistic $t_{\alpha/2}$ is 6.678155

(d) The p value is 1.2×10⁻¹¹

(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high

Step-by-step explanation:

(a) Here we have the data as follows;

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The median is = 1478

Therefore we have above the median

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484

The mean, $\bar{x}_{1}$ = 1541.9

Standard deviation, σ₁ = 41.38224257

n₁ = 10

Below the median

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The mean, $\bar{x}_{2}}$ = 1433.9

Standard deviation, σ₂ = 30.04812806

n₂ = 10

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

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$t_{\alpha/2} =\frac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}$

df = 10 - 1 = 9, α = 0.05

Therefore, the test statistic $t_{\alpha/2}$ = 6.678155

(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹

Critical z at 5% confidence level = 1.645

Since P << 0.05, e reject the

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