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Serjik [45]
3 years ago
11

Jared can read 8 pages in 15 minutes. A this rate, how many pages can he read in 45 minutes?

Mathematics
2 answers:
fenix001 [56]3 years ago
5 0

Answer:

24 pages

Step-by-step explanation:

8:15 = a:45

Cross multiply:

a × 15 = 8 × 45

15a = 8 × 45

15a = 360

15a ÷ 15 = 360 ÷ 15

a = 360 ÷ 15

a = 24

24 pages

Alona [7]3 years ago
3 0

Answer:

...

Step-by-step explanation:

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14 divided 3/11 +14 x
slamgirl [31]

Answer:

14x+14/33

Step-by-step explanation:

14÷3/11+14x=

5 0
4 years ago
Write 2/3 as a decimal.
Mekhanik [1.2K]

Answer:

0.67

Step-by-step explanation:

2/3 = 0.6666

0.6666 rounded to the nearest hundredth = 0.67

3 0
3 years ago
To the nearest hundredth, what is the length of line segment AB?
Vesnalui [34]

Answer:

5.83 units

Step-by-step explanation:

See attached picture.  You can use the Pythagorean Theorem.  I added some lines to show the legs of a right triangle.  AB is the hypotenuse.

3^2+5^2=(AB)^2\\\\9+25 = (AB)^2\\\\34=(AB)^2\\\\AB=\sqrt{34} \approx 5.83

5 0
2 years ago
Write the expanded form for this expression <br><br> -1/2(y - x)
grin007 [14]

Answer:  =-1/2x+-1/2y

Step-by-step explanation:

5 0
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
12345 [234]

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})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

6 0
3 years ago
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