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2 years ago

8

A batch of one hundred cars is inspected by testing four randomly selected cars. If one of the four is defective, the batch is r

ejected. What is the probability that the batch is accepted if it contains five defective cars?

1 answer:

Y_Kistochka [10]2 years ago

3 0

Answer: The probability that the batch is accepted is 0.812.

Step-by-step explanation:

Since we have given that

Number of defective cars = 5

Total number of cars = 100

Probability of getting first car non defective = $\dfrac{95}{100}$

probability of getting second car non defective = $\dfrac{94}{99}$

Probability of getting third car non defective = $\dfrac{93}{98}$

Probability of getting fourth car non defective = $\dfrac{92}{97}$

Using the multiplication rule, probability that the batch is accepted is given by

$\dfrac{95}{100}\times \dfrac{94}{99}\times \dfrac{93}{98}\times \dfrac{92}{97}\\\\=0.812$

Hence, the probability that the batch is accepted is 0.812.

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Answer:

y=x+1

Step-by-step explanation:

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2 years ago

A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point est

xxTIMURxx [149]

Answer:

Point estimate for the population variance = 3.92 * $10^{-3}$ .

Step-by-step explanation:

We are given that a sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters ;

X X - $Xbar$ $(X-Xbar)^{2}$

1.13 1.13 - 1.188 = -0.058 3.364 * $10^{-3}$

1.15 1.15 - 1.188 = -0.038 1.444 * $10^{-3}$

1.15 1.15 - 1.188 = -0.038 1.444 * $10^{-3}$

1.24 1.24 - 1.188 = 0.052 2.704 * $10^{-3}$

1.27 1.27 - 1.188 = 0.082 <u> 6.724 * </u> $10^{-3}$ <u> </u>

$\sum (X-Xbar)^{2}$ <u>= 0.01568 </u>

Firstly, Mean of above data, $Xbar$ = $\frac{\sum X}{n}$ = $\frac{1.15+1.24+1.15+1.27+1.13}{5}$ = 1.188

Point estimate of Population Variance = Sample variance

= $\frac{\sum (X-Xbar)^{2}}{n-1}$ = $\frac{0.01568}{4}$ = 3.92 * $10^{-3}$ .

Therefore, point estimate for the population variance = 3.92 * $10^{-3}$ .

5 0

2 years ago

Read 2 more answers

Simplify 7y+2x - 10y+3x to the second power -10x

lara [203]

$7y+2x - 10y+3x^2-10x=(3x^2)+(2x-10x)+(7y-10y)\\\\=3x^2-8x-3y$

4 0

2 years ago

(Q7) Determine the graph of the polar equation r = 12/6+4 cos theta.

masya89 [10]

ANSWER

b.

EXPLANATION

The given polar equation is,

$r = \frac{12}{6 + 4 \cos( \theta) }$

Cross multiply to get,

$r(6 + 4 \cos( \theta) ) = 12$

$6r + 4r\cos( \theta) ) = 12$

Substitute,

$r\cos( \theta) = x$

and

$r = \sqrt{ {x}^{2} + {y}^{2} }$

$6 \sqrt{ {x}^{2} + {y}^{2} } + 4x = 12$

$3\sqrt{ {x}^{2} + {y}^{2} } + 2x = 6$

$3\sqrt{ {x}^{2} + {y}^{2} } = 4 - 2x$

$9({x}^{2} + {y}^{2}) =( 4 - 2x ) ^{2}$

$9{x}^{2} + 9 {y}^{2} =16 - 16x + 4x ^{2}$

$5{x}^{2} + 9 {y}^{2} + 16x = 16$

This is an ellipse whose center is shifted to the left with orientation on the x-axis

The correct choice is B.

8 0

3 years ago

One vegetable warehouse had 210 tons of potatoes, another one had 180 tons. 90 tons of potatoes were delivered to the first and

kow [346]

Answer:

The number of days in which the amount in warehouse #2 is 1.2 greater than warehouse #1 is 6 days.

Step-by-step explanation:

Let

x: number of days

y: amount of potatoes

We can express the amount of potatoes in each warehouse using linear equations:

Warehouse #1: $y_1 = 90x+210$

Warehouse #2: $y_2 = 120x+180$

We need to find the value of x when $1.2*y_{1} =y_{2}$

Substituting the expressions for y1 and y2 in $1.2*y_{1} =y_{2}$ :

$1.2*(90x+210) =120x+180$

Applying distributive law:

$108x+252 =120x+180$

Solving the equation by subtracting 108x and 180 on both sides:

$108x+252-108x-180 =120x+180-108x-180$

$252-180 =120x-108x$

$72 =12x$

Now we divide by 12 on both sides:

$\frac{72}{12} =\frac{12x}{12}\\6=x\\x=6$

Therefore, the number of days in which the amount in warehouse #2 is 1.2 greater than warehouse #1 is 6 days.

4 0

3 years ago