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Rzqust [24]
3 years ago
11

Find the number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4.

Mathematics
2 answers:
Katen [24]3 years ago
7 0

Answer:

1364

Step-by-step explanation:

Paraphin [41]3 years ago
4 0

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

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If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c
dolphi86 [110]

Given:

(ax+2)(bx+7)=15x^2+cx+14

And

a+b=8

Required:

To find the two possible values of c.

Explanation:

Consider

\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}

So

\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}

And also given

a+b=8---(2)

Now from (1) and (2), we get

\begin{gathered} a+\frac{15}{a}=8 \\  \\ a^2+15=8a \\  \\ a^2-8a+15=0 \end{gathered}a=3,5

Now put a in (1) we get

\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}

When a=5 and b=3, we have

\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}

Final Answer:

The option D is correct.

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luda_lava [24]

Answer:

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Step-by-step explanation:

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The confidence interval is directly proportional to the confidence interval.

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On increasing the confidence level the confidence level widens. And on decreasing the confidence level the confidence level gets narrower.

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Step-by-step explanation:

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3 0
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Fynjy0 [20]
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____________________

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Average rate of change: 
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