24.6 so I’m just spamming to make it 20 characters cause i really need points for answers but that said incorrect answer so let’s continue it’s definitely b) 113.01ft whilst c) being 116.46 ft.... : )

5 0

3 years ago

You go to the carnival Wingate $10 to spend on Gameday rise each Ricossa dollar in each game cost $.75 write a linear equation i

m_a_m_a [10]

Answer:

Step-by-step explanation:

you divide 10 by 0.75

4 0

3 years ago

The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 1100lbs.

gizmo_the_mogwai [7]

Answer:

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(920≤ x≤1730) = 0.7078

Step-by-step explanation:

Step(i):-

Given mean of the Population = 1100 lbs

Standard deviation of the Population = 300 lbs

Let 'X' be the random variable in Normal distribution

Let x₁ = 920

$Z = \frac{x-mean}{S.D} = \frac{920-1100}{300} = - 0.6$

Let x₂ = 1730

$Z = \frac{x-mean}{S.D} = \frac{1730-1100}{300} = 2.1$

Step(ii)

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(x₁≤ x≤x₂) = P(Z₁≤ Z≤ Z₂)

= P(-0.6 ≤Z≤2.1)

= P(Z≤2.1) - P(Z≤-0.6)

= 0.5 + A(2.1) - (0.5 - A(-0.6)

= A(2.1) +A(0.6) (∵A(-0.6) = A(0.6)

= 0.4821 + 0.2257

= 0.7078

Conclusion:-

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(920≤ x≤1730) = 0.7078

5 0

3 years ago

Sam read 3/8 of a book in 3/4 of an hour. At this rate, how long will it take him to read the whole book?

avanturin [10]

2 hours he is reading 1/8 of the book in 1/4 of an hour 1/4x8=2

5 0

3 years ago

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