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2 years ago

6

The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 900lbs.

Find the probability that the weight of a randomly selected steer is between 1220 and 1320lbs. Round your answer to four decimal places.

1 answer:

katen-ka-za [31]2 years ago

3 0

Answer:

0.0369

Step-by-step explanation:

normalcdf (1220,1320,900,200) is 0.0369

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Solve x^2 x=0.factorise.

Semenov [28]

$x^2\pm x=0\\\\x(x\pm1)=0\iff x=0\ or\ x\pm1=0\\\\\boxed{x=0\ or\ x=\mp1}\\\\if\ x^2+x=0\ then\ x=0\ or\ x=-1\\if\ x^2-x=0\ then\ x=0\ or\ x=1$

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3 years ago

Write the equation of the line that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5)

Sedbober [7]

Answer: $\bold{y=-\dfrac{1}{3}x-1}$

<u>Step-by-step explanation:</u>

(4, 1) & (2, -5)

First, find the slope (m) and then the perpendicular (opposite reciprocal) slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\\m=\dfrac{-5-1}{2-4} = \dfrac{-6}{-2}=3\quad \rightarrow \quad m_{\perp}=-\dfrac{1}{3}\\$

Next, find the midpoint of (4, 1) and (2, -5):

$Midpoint=\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{4+2}{2},\dfrac{1-5}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{6}{2},\dfrac{-4}{2}\bigg)\\\\\\.\qquad \qquad=(3, -2)$

Lastly, input the perpendicular slope and the midpoint into the Point-Slope formula to find the equation of the line:

$y - y_1 = m_{\perp}(x - x_1)\\\\y - (-2) = -\dfrac{1}{3}(x - 3)\\\\y + 2=-\dfrac{1}{3}x +1\\\\y =-\dfrac{1}{3}x - 1\\$

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3 years ago

Can someone help me plz

wolverine [178]

Answer:

75°

Step-by-step explanation:

Recall: SOH CAH TOA

Reference angle = ? = θ

Side length opposite to reference angle = 27

Hypotenuse length = 28

Apply SOH, which is:

$sin(\theta) = \frac{Opposite}{Hypotenuse}$

Substitute

$sin(\theta) = \frac{27}{28}$

$\theta = sin^{-1}(\frac{27}{28})$

$\theta = 75 degrees$ (nearest degree)

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3 years ago

Write the equation of the line fully simplified slope-intercept form.

Vladimir [108]

Y=-2/3x-5
i think is the answer

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3 years ago

When mrs. myles gave a test, the scores were normally distributed with a mean of 72 and a standard deviation of 8. this means th

irina [24]

The empirical rule states that in a normal distribution,
68% of data is within 1 std deviation of the mean
95% of data is within 2 std deviation of the mean
99.7% of data is within 3 std deviation of the mean

In this case 95% of the cases would be within two std deviations of the mean
mean - 8 and mean + 8
72 - 8 = 64 and 72 + 8 = 80

then 95% of the scores are between 64% and 80% on the test.

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3 years ago