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tester [92]
3 years ago
10

PLS HELP ME I HAVE ONLY 10 MINS

Mathematics
1 answer:
anastassius [24]3 years ago
8 0
Sorry habibti it’s okay tho f school
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Triangle ABC is congruent to triangle DEF. If DE = 41, EF = 23, DF = 36, and AB = 8x − 7, what is x? Round to the nearest tenth.
Nadya [2.5K]

The side AB is equal to the side DE. Then the value of x is 6.

<h3>What is the triangle?</h3>

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

Triangle ABC is congruent to triangle DEF.

If DE = 41, EF = 23, DF = 36, and AB = 8x − 7.

Then the side AB is equal to the side DE.

    AB = DE

8x - 7 = 41

    8x = 48

      x = 6

Then the value of x is 6.

More about the triangle link is given below.

brainly.com/question/25813512

#SPJ1

5 0
2 years ago
Read 2 more answers
Find the 7th term of the sequence 1,5/4,25/16
lutik1710 [3]

9514 1404 393

Answer:

  15625/4096

Step-by-step explanation:

The sequence is geometric with first term 1 and common ratio 5/4. The general term is ...

  an = a1·r^(n-1)

  an = 1·(5/4)^(n-1)

Then the 7th term is ...

  a7 = (5/4)^(7-1) = 15625/4096

7 0
3 years ago
The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000
valina [46]

Answer: 61,750 miles

Step-by-step explanation:

Given : The p-value of the tires to outlast the warranty = 0.96

The probability that corresponds to 0.96 from a Normal distribution table is 1.75.

Mean : \mu=60,000\text{ miles}

Standard deviation : \sigma=1000\text{ miles}

The formula for z-score is given by  : -

z=\dfrac{x-\mu}{\sigma}\\\\\Rightarrow\ 1.75=\dfrac{x-60000}{1000}\\\\\Rightarrow\ x-60000=1750\\\\\Rightarrow\ x=61750

Hence, the tread life of tire should be 61,750 miles if they want 96% of the tires to outlast the warranty.

3 0
3 years ago
The Campers in cabin D walked 1/6 of the way to Mammoth Mountain. how many miles did the walk ?
svet-max [94.6K]
What is the question ?
4 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
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})

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

P(at least 1 dull blade)=0.7068

5 0
3 years ago
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