Here's number 5. If you need more explanation just let me know.
22
1 answer:
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Answer:
sin-¹(10/11)
Step-by-step explanation:
The height of the building is 100 ft and the ladder is 110ft long .We can imagine this situation as a right angled ∆. For figure refer to the attachment .
- We can use ratio sine here. Let the angle of elevation be theta.
<u>In </u><u>∆</u><u> </u><u>ABC </u><u>:</u><u>-</u><u> </u>
=> sinθ = p/h
=> sinθ = 100ft / 110 ft
=> sinθ = 10/11
=> θ = sin -¹ ( 10/11) .
<h3><u>Hence </u><u>the</u><u> </u><u>angle</u><u> of</u><u> elevation</u><u> </u><u>is </u><u>sin </u><u>-</u><u>¹</u><u> </u><u>(</u><u> </u><u>1</u><u>0</u><u>/</u><u>1</u><u>1</u><u>)</u><u>. </u><u> </u></h3>
To find the z-score for a weight of 196 oz., use
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.