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Paul [167]
2 years ago
9

Mason used distributive property of multiplication to help him solve the equation 3 x 7 = ________. Mason did not get the same a

nswer for each side of the expression. Explain the steps required to correct his mistake.
Mathematics
1 answer:
cupoosta [38]2 years ago
3 0

Answer:

See below

Step-by-step explanation:

3*7

3(3+4)

3*3 + 3*4

9 + 12

21

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More help please!!!!!!!!!!!!!!!!!!!!!!!
Gemiola [76]

Answer:

3 1/3

Step-by-step explanation:

10+z=z+z+z+z

10+z=4z

10=3z

z=3 1/3

8 0
3 years ago
Hey can someone help:)?​
Zolol [24]
I think it’s the first one
7 0
2 years ago
Examine the table below showing the merchandising sales and answer the questions that follow.
-BARSIC- [3]

Answer:

The mean amount spent = £14.80

The modal product is hoodies.

Step-by-step explanation:

Here

25 T shirts each costing £10 were purchased

Amount spent on T shirts = 25 × 10 = £250

30 key ring of £5 were purchased

Amount spent on key ring = 30 × 5 = £150

40 Hoodies for £25 each

Amount spent on Hoodies = 40 × 25 =£1000

30 CD's for £15 each

Amount spent on CD's = 30 ×15 = £450

Total amount spent = 250 +150 + 1000 + 450 = £1850

Total items purchased = 25 + 30 + 40 + 30 = 125

Mean amount spent = \frac{1850}{125}= 14.80

The modal product is Hoodies as the maximum number of hoodies were purchased.


6 0
3 years ago
Nine less than twice a number is 49.<br> What is the number?
ratelena [41]

Answer:

29

Step-by-step explanation:

Assume the number as x

2x - 9 = 49

2x = 58

x = 29

7 0
2 years ago
A manufacturer of matches randomly and independently puts 23 matches in each box of matches produced. The company knows that one
d1i1m1o1n [39]

Answer:

0.9855 or 98.55%.

Step-by-step explanation:

The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:

P(X\leq 1)=P(X=0)+P(X=1)\\P(X\leq 1)=(1-p)^{23}+23*(1-p)^{23-1}*p\\P(X\leq 1)=(1-0.008)^{23}+23*(1-0.008)^{23-1}*0.008\\P(X\leq 1)=0.8313+0.1542\\P(X\leq 1)=0.9855

The probability  that a matchbox will have one or fewer matches with a flaw is 0.9855 or 98.55%.

8 0
3 years ago
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