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Kazeer [188]
3 years ago
12

When is the substitution method a better method than graphing for solving a system of

Mathematics
1 answer:
baherus [9]3 years ago
8 0

Answer:

Frankly, substitution is almost always a better method. Graphing is a very precise and time consuming process where you have to calculate several..

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Can someone help me not sure how to do it
tatyana61 [14]
First add the perimeter
3.65m + 2.24m + 2.24m + 3.65m + 2.57m = 14.35 m
then convert meters to centimeters (just move the decimal place two places to the right) 
answer is 1,435 cm

6 0
3 years ago
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Tyshdhhehdrhrhfhfurjrjd
jek_recluse [69]

Answer:

.....

Step-by-step explanation:

7 0
3 years ago
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Sherman has two sequences . The first sequence is described by the explicit rule f(n) = 15n + 4 and the second sequence is descr
Viktor [21]

Answer:

Step-by-step explanation:

Given the explicit function as

f(n) = 15n+4

The first term of the sequence is at when n= 1

f(1) = 15(1)+4

f(1) = 19

a = 19

Common difference d = f(2)-f(1)

f(2) = 15(2)+4

f(2) = 34

d = 34-19

d = 15

Sum of nth term of an AP = n/2{2a+(n-1)d}

S20 = 20/2{2(19)+(20-1)15)

S20 = 10(38+19(15))

S20 = 10(38+285)

S20 = 10(323)

S20 = 3230.

Sum of the 20th term is 3230

For the explicit function

f(n) = 4n+15

f(1) = 4(1)+15

f(1) = 19

a = 19

Common difference d = f(2)-f(1)

f(2) = 4(2)+15

f(2) = 23

d = 23-19

d = 4

Sum of nth term of an AP = n/2{2a+(n-1)d}

S20 = 20/2{2(19)+(20-1)4)

S20 = 10(38+19(4))

S20 = 10(38+76)

S20 = 10(114)

S20 = 1140

Sum of the 20th terms is 1140

7 0
2 years ago
Hector entered a 3-day bike race. He traveled 20t miles on the first day, 15(t+6) miles the second day, and 25(t-3) miles the th
madreJ [45]
20t +15t+90+25t-75=60t+15
60t+15/3=20t+5 as the average
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8 0
3 years ago
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Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. ​(a) Descri
OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

\mu = 74

and

\sigma = 6

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

\mu = 74

The standard error of the means becomes the standard deviation of the sampling distribution.

\sigma_ { \bar X }  =  \frac{ \sigma}{ \sqrt{n} }  \\ \sigma_ { \bar X }  =  \frac{ 6}{ \sqrt{36} }  = 1

Part b) We want to find

P(\bar X \:>\:75.9)

We need to convert to z-score.

P(\bar X \:>\:75.9)  = P(z \:>\: \frac{75.9 - 74}{1} )  \\  = P(z \:>\: \frac{75.9 - 74}{1} ) \\  = P(z \:>\: 1.9) \\  = 0.0287

Part c)

We want to find

P(\bar X \: < \:71.95)

We convert to z-score and use the normal distribution table to find the corresponding area.

P(\bar X \: < \:71.95)  = P(z \: < \: \frac{71.9 5- 74}{1} )  \\  = P(z \: < \: \frac{71.9 5- 74}{1} ) \\  = P(z \: < \:  - 2.05) \\  = 0.0202

Part d)

We want to find :

P(73\:

We convert to z-scores and again use the standard normal distribution table.

P( \frac{73 - 74}{1} \:< \: z

5 0
3 years ago
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