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3 years ago

Need help on test Angles and triangles 2

Mathematics

1 answer:

erma4kov [3.2K]3 years ago

6 0

Alternate interior: <2 and <7
(They are on opposite sides of the transversal line, but inside the parallel lines.)

Corresponding: <1 and <3
(They have the same measure because they are in the same spot above the parallel line.)

Alt exterior: <1 and <8
(They are on opposite sides of the transversal, but on the outside of the parallel lines.)

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Philip is conducting a survey to determine the weight of 100 cows on a farm. Which of the following questions is an appropriate

liubo4ka [24]

Because Philip is conducting a survey that pertains to the weight of the cows, he needs statistics that directly relate to weight.

The cows being overweight doesn't give Philip concrete information on the weight of the cows, as it doesn't give the amount of pounds a cow weighs. Average age has nothing to do with weight, and normal body weight doesn't give concrete information on weight.

The answer is "<span>What is the weight of each cow on the farm?".</span>

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3 years ago

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When ringing up a customers cashier needs 25 seconds to process payment as well as 1 second to scan each item being purchased. H

Sergio [31]

Multiply the amount of seconds to ring up items by the amount of items.

1 × 17 = 17

Now add the amount of seconds to process a payment to the product.

17 + 25 = 42

<h2>Answer:</h2>

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3 years ago

The width of a rectangle is the length minus 2 units. the area of the rectangle is 35 units

Mariana [72]

Answer:

$Length=7units\\\\Width=5units$

Step-by-step explanation:

Width=(Length-2) units

W=(L-2) units

Area of the rectangle= 35 square units

Area = length* Width

$35= L*(L-2)\\\\35=L^2-2L$

Subtracting 35 both sides:

$L^2-2L-35=0$

Solving the quadratic equation for 'L' ;

Using factorization:

$L^2+5L-7L-35=0$

Taking common from the equation :

$L(L+5)-7(L+5)=0\\\\(L+5)(L-7)=0\\\\L+5=0\\\\L=-5$

$L-7=0\\\\L=7$

The length cannot be negative, therefore Length(L)= 7

$Width=Length-2\\\\=7-2\\\\=5 units$

$Length=7units\\\\Width=5units$

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3 years ago

a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade

son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

$P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})$