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3 years ago

NEED HELP ILL MARK BRAINLIEST

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

6 0

I’ve attached my work
b is your answer
I just drew a sketch of a graph and saw if it would be a line

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Help me lawd!!!!!!!jkdmfkfm

Klio2033 [76]

This is equivalent to:

(2.2533/2.59)(10^8/10^4)

(0.87)(10^4) which is:

0.87X10^4 which is equal to:

0.87X10000 which is equal to:

8.7X1000 and since 1000=10^3 we can say:

8.7X10^3

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3 years ago

Please help will give brainliest

natulia [17]

Answer:

A) it wouldn't fit

Step-by-step explanation:

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3 years ago

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No links pls

Jobisdone [24]

Answer: D: the point (0,0) contains the x-intercept

Step-by-step explanation:

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3 years ago

Help please I don't know this I'm a 5th grader. help please anybody

Arte-miy333 [17]

He switched the 5 and 6 into different places

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4 years ago

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If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 477 viruses would d

melisa1 [442]

Answer:

0.01596

Step-by-step explanation:

A scientist claims that 8% of the viruses are airborne

Given that:

The population proportion p = 8%

The sample size = 477

We can calculate the standard deviation of the population proportion by using the formula:

$\sigma_p = \sqrt{\dfrac{p(1-p)}{n}}$

$\sigma_p = \sqrt{\dfrac{0.8(1-0.8)}{477}}$

$\sigma_p = \sqrt{\dfrac{0.0736}{477}}$

$\sigma_p = 0.02098$

The required probability can be calculated as:

$P(| \hat p - p| > 0.03) = P(\hat p - p< -0.03 \ or \ \hat p - p > 0.03)$

$= P \bigg ( \dfrac{\hat p -p }{\sqrt{\dfrac{p(1-p)}{n}}} < -\dfrac{0.03}{0.0124} \bigg ) + P \bigg ( \dfrac{\hat p -p }{\sqrt{\dfrac{p(1-p)}{n}}} >\dfrac{0.03}{0.0124} \bigg )$

= P(Z < -2.41) + P(Z > 2.41)

= P(Z < -2.41) + P(Z < -2.41)

= 2P( Z< - 2.41)

From the Z-tables;

$P(| \hat p - p| > 0.03)$ = 2 ( 0.00798

$P(| \hat p - p| > 0.03)$ = 0.01596

Thus, the required probability = 0.01596

3 0

3 years ago