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4 years ago

Between two tenths of a mile will the second relay exchange occur? Explain.

Mathematics

2 answers:

suter [353]4 years ago

7 0

2/10

There is one tenth in a mile. I have the same question. But I am just guessing.

kenny6666 [7]4 years ago

4 0

Yes it will occur because two tenths of a mile the second relay will exchange

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Tomas rides his bike at a steady rate of 4 miles every 6 minutes. Find the unit rate of this proportional relationship

ddd [48]

4 miles every 6 minutes
So proportional relationship is 4:6
Simplified its 2:4

6 0

4 years ago

This is what it says

Schach [20]

3x+2=-x+10

3x+2+x=-x+10+x

4x+2=10

4x+8

4x/4=8/4

x=2

The solution set is {2}

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3 years ago

In a test of a printed circuit board using a random test pattern, an array of 16 bits is equally likely to be 0 or 1. Assume the

Doss [256]

Answer:

A) b) and c) answer has been explained below.

Step-by-step explanation:

For the question would use 1's and 0's and taking probabilities to be equal for both using random test pattern whose formula is when p = q

then Simplify to

P[k] = nCk /2^n

A.Probability that all bits are 1s

16c16/2^16 = 1/65536

B. Probability that all bits are 0s

16c0/2^16 = 1/65536

C.the probability that exactly 8 bits are 1s and 8 bits are 0s

16c8/2^16 = 12870/65536 =>0.1963 ≈ 19.63%

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4 years ago

The following random sample from a population whose values were normally

vodomira [7]

Answer:

f urself

Step-by-step explanation:

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3 years ago

When the components arrive, the company selects a random sample from the shipment and subjects the selected components to a rigo

Ghella [55]

Answer:

(0.059, 0.133)

Step-by-step explanation:

Sample size = n = 250

Number of units which failed the test = x = 24

Proportion of units which failed the test = $\frac{x}{n} = \frac{24}{250} =\frac{12}{125}$ = 0.096

Proportion of units which did not fail the test = q = 1 - p = 1 - 0.096 = 0.904

Confidence level = 95%

z-value for the confidence level = z = 1.96

The true proportion of the components that fail to meet the specification would be:

$(p-z\sqrt{\frac{p \times q}{n}} , p+z\sqrt{\frac{p \times q}{n}})$

Using the values, we get:

$(0.096-1.96 \times \sqrt{\frac{0.096 \times 0.904}{250}} , 0.096+1.96 \times \sqrt{\frac{0.096 \times 0.904}{250}})\\\\ =(0.059,0.133)$

Thus, 95% confidence interval estimate for the true proportion of components, p, that fail to meet the specifications is (0.059, 0.133)

4 0

3 years ago