Mackenzie read 64 pages in 8 days. She read the same number of pages each day. How many pages did she read on Tuesday?

Mathematics

1 answer:

Gekata [30.6K]3 years ago

7 0

Answer: 8

Step-by-step explanation:

You have to divide 64 by 8 which means she reads 8 pages a day

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How do I do this Q11

SashulF [63]

Look at one of the vertices of the heptagon where two squares meet. The angles within the squares are both of measure 90 degrees, so together they make up 180 degrees.

All the angles at one vertex must clearly add up to 360 degrees. If the angles from the squares contribute a total of 180 degrees, then the two remaining angles (the interior angle of the heptagon and the marked angle) must also be supplementary and add to 180 degrees. This means we can treat the marked angles as exterior angles to the corresponding interior angle.

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a 20 ft piece of string is cut into two pieces so that the longer piece is 5 feet longer than twice the shorter piece. If the sh

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Longer piece is 15 and shorter piece is 5

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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)

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