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creativ13 [48]
3 years ago
13

a bus driver drives 547.25 miles on day 1 , on day 2 , he drives 327.875 , how many more miles did he drive the first day than t

he second day
Mathematics
1 answer:
quester [9]3 years ago
3 0
He drove 246.375 more miles the second day
You might be interested in
(CO 3) On average, the parts from a supplier have a mean of 97.5 inches and a standard deviation of 6.1 inches. Find the probabi
ololo11 [35]

Answer:

C. Probability is 0.90, which is inconsistent with the Empirical Rule.

Step-by-step explanation:

We have been given that on average, the parts from a supplier have a mean of 97.5 inches and a standard deviation of 6.1 inches.

First of all, we will find z-score corresponding to 87.5 and 107.5 respectively as:

z=\frac{x-\mu}{\sigma}

z=\frac{87.5-97.5}{6.1}

z=\frac{-10}{6.1}

z=-1.6393

z\approx-1.64

z=\frac{x-\mu}{\sigma}

z=\frac{107.5-97.5}{6.1}

z=\frac{10}{6.1}

z=1.6393

z\approx 1.64

Now, we need to find the probability P(-1.64.

Using property P(a, we will get:

P(-1.64

From normal distribution table, we will get:

P(-1.64

P(-1.64

P(-1.64

Since the probability is 0.90, which is inconsistent with the Empirical Rule, therefore, option C is the correct choice.

6 0
3 years ago
Two numbers which differ by 46 using 29,17,38,71,35,63,24
Nikitich [7]

I think 24 but im not 100 percent sure more like 60               .

7 0
3 years ago
Ten percent of the engines manufactured on an assembly line are defective.
antoniya [11.8K]

Answer:

a) the probability that the first non-defective engine will be found on the second trial is 0.09

b) probability that the third non-defective engine will be found on the fifth trial is 0.00486

c) the Mean is 1.1111  and Variance is 0.1235  

d) the Mean is 3.3333and Variance is 0.3704  

Step-by-step explanation:

Firstly;

Let x be the number of trail on which the rth defective occurs

Also let the probability that an occurrence of a defective be p = 0.10

Here, x follows negative binomial distribution with parameters r and p

The probability mass function of X is as follows:

P(X = x) = [ x -1  p^r ( 1 - p)^(x-r)       ; x = r, r + 1, r + 2, ...

                 r - 1 ]

=   [ x -1  (0.10)^r ( 1 - 0.10)^(x-r)

     r - 1 ]

= [ x -1  (0.10)^r ( 0.9)^(x-r)  ..........n 1..... let this be equatio

    r - 1 ]

This represents the probability that the rth success occurs on the xth trail.  

a)

probability that the first non-defective engine will be found on the second trial?

Substitute r = 1 and x = 2 in equation 1  

P(X = 2)  = [ 2 - 1   (0.10)¹ ( 0.90 )²⁻¹

                   1 - 1 ]

= (0.10) × (0.90)

= 0.09

Therefore, the probability that the first non-defective engine will be found on the second trial is 0.09

b)

probability that the third non-defective engine will be found on the fifth trial?

So we substitute r = 3 and x = 5 in equation 1  

P(X = 5) = [ 5 - 1  (0.10)³ ( 0.90)⁵⁻³

                  3 - 1 ]

=    [ 4    (0.10)³ ( 0.9)²

      2 ]

= 0.00486

Therefore, probability that the third non-defective engine will be found on the fifth trial is 0.00486

Now the formula for the mean of the negative binomial distribution is as follows:  

Mean u = r / p    ------- let this be equation 2

The formula for the variance of the negative binomial distribution also is as follows:  

Variance α² = rq / p²   ---------- let this be equation 3

so

c)

the mean and variance of the number of trials on which the first non-defective engine is found.  

First, let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10  

Now we substitute r = 1, p = 0.90 and q = 0.10 in equation 2 & 3simultaneously,

the mean and variances are as follows;

Mean = r/p = 1/0.90 = 1.1111

Variance = rq/p² = (1)(0.10) / (0.90)² = 0.1235  

Therefore the Mean is 1.1111  and Variance is 0.1235  

d)

the mean and variance of the number of trials on which the third non-defective engine is found  

Let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10

Now we substitute r = 3, p = 0.90 and q = 0.10 in equation (2) and (3) simultaneously,

the mean and variances are;

Mean = r/p = 3/0.90 = 3.3333

Variance = rq/p² = (3)(0.10) / (0.90)² = 0.3704

Therefore the Mean is 3.3333 and Variance is 0.3704    

5 0
3 years ago
Does anyone know the last two? Match the correct property to the equation showing that property.
sergij07 [2.7K]

Explanation: The 4th one is the distributive property because it shows an     equal sign and the 5th one is the inverse property of addition because it uses more than addition.

Answer: 4th one is distributive property 5th one is inverse property of addition

P.S. I am sure 95% sure that I am correct.

8 0
1 year ago
What is the value of -8 (17-12) over -2(8-(-2))<br><br> A. -4<br> B. -2<br> C. 2 <br> D. 4
Ivan

Answer:

C=2

Step-by-step explanation:

-8(17-12)= -8(5)= -40

then

-2(8-(-2))= -2(10) = -20

so then

-40/-20

=-2

3 0
3 years ago
Read 2 more answers
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