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aleksandr82 [10.1K]
3 years ago
11

Mikhael wanted to rewrite the conversion factor "1 yard ≈ 0.914 meter" to create a conversion factor to convert meters to yards.

Complete the explanation to tell how Mikhael should finish his conversion. If necessary, round to three decimal places.
(select)
1 yard by 0.914 m gives the rate of yards per meter. So 1 meter ≈
yards.
plz need help
Mathematics
1 answer:
Nataly [62]3 years ago
6 0

Answer:

1\ meter = 1.094\ yard

Step-by-step explanation:

Given

1\ yard = 0.914\ meter

Required

Create a convert meter to yard expression

1\ yard = 0.914\ meter

Divide both sides by 0.914

\frac{1\ yard}{0.914} = \frac{0.914\ meter}{0.914}

\frac{1\ yard}{0.914} = 1\ meter

1.09409190372\ yard = 1\ meter

1\ meter = 1.09409190372\ yard

1\ meter = 1.094\ yard --- approximated

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A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556
MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

The formula to find the standard deviation:-

\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}

The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.

Then,  \overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560

Now, \sum_{i=1}^{10}(x_i-\overline{x})^2=0^2+2^2+(-4)^2+(-2)^2+0^2+(-4)^2+(-1)^2+1^2+5^2+3^2\\\\\Rightarrow\ \sum_{i=1}^{10}(x_i-\overline{x})^2=76

Then, \sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76

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6 0
3 years ago
Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi
ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles  carry just one person.    

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

\mu = np = 92(0.64) = 58.88

\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60

We have to evaluate the probability that more than 47 cars carry just one person.

P(x \geq 47)

After continuity correction, we will evaluate

P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)

= 1 - P(z \leq -2.6913)

Calculation the value from standard normal z table, we have,  

P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0
3 years ago
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