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galina1969 [7]
3 years ago
15

Evaluate the expression for x = 2, y = -3, and z = -5. * 2x + 3y + z Your answer

Mathematics
2 answers:
Lunna [17]3 years ago
6 0
Plug in

2(2) + 3(-3) + (-5)
4 + (-9) + (-5)
4 + (-14)
-10

Answer is -10
2x + 3y + z = -10
algol133 years ago
6 0

Answer:

-10

Step-by-step explanation:

replace each letter with each number given, like shown below.

2(2)+3(-3)+(-5)

= 4+-9+-5

= -5+-5

= -10

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The answer to that is -15-36 hope that helps you with your problem.
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An ANOVA procedure is used for data obtained from four populations. Four samples, each comprised of 30 observations, were taken
tatiyna

Answer:

D. 3 and 116

Step-by-step explanation:

d.f.N = k - 1 (numerator degrees of freedom) = 4 - 1 = 3

N = 4 × 30 = 120

d.f.D = N - k (denominator degrees of freedom) = 120 - 4 =116

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The Eiffel Tower is about 301 meters tall. Cooper is making a scale model of it using the scale 750 meters : 1 meter.
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102·104 simiply<br> please i need this only answer to pass my class its homework please
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Hope it helped
4 0
3 years ago
Find the number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4.
Paraphin [41]

As we can see that there are 6 digits in 100,000 and its is the smallest number we can have in 6 digit. So all numbers less than 100,000 will be 1-digit, 2-digits, 3-digits, 4-digits and 5-digits numbers made from 1,2,3,4 with repetitions allowed.

Case 1: All 1 -digit numbers

We will have numbers 1,2,3,4. So total 4 integers for this one

Case2: All 2-digit numbers

We can fill 1 digit place in 4 ways ( can choose any number out of 1,2,3,4). Then again we can fill 2nd digit place in 4 ways ( can choose any number out of 1,2,3,4). So all together we will have 4 × 4 = 16 integers for this one

Case3: All 3-digits numbers

We can fill 1 digit place in 4 ways ( can choose any number out of 1,2,3,4). Then again we can fill 2nd digit place in 4 ways ( can choose any number out of 1,2,3,4). similarly we can fill 3rd digit place in 4 ways (again any number out of 1,2,3,4). So all together we will have 4 × 4 × 4 = 64 integers for this one.

Case4: All 4-digit numbers

Again we can fill 1st digit place in 4 ways, then 2nd digit place in 4 ways, 3rd digit place in 4 ways, 4th digit place in 4 ways. So all together there will be 4 × 4 × 4 × 4 = 256 integers for this one.

Case5: All 5-digit numbers

Again we can fill 1st digit place in 4 ways, then 2nd digit place in 4 ways, 3rd digit place in 4 ways, 4th digit place in 4 ways, 5th digit place also in 4 ways. So all together there will be 4 × 4 × 4 × 4 × 4 = 1024 integers for this one

Adding results of all 5 cases we get,

Total integers = 4 + 16 +64 + 256 + 1024 = 1364 integers.

So thats the final answer

4 0
3 years ago
Read 2 more answers
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