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tamaranim1 [39]
3 years ago
15

Graph the inequality on a number line m<1

Mathematics
1 answer:
BARSIC [14]3 years ago
6 0

Answer:

In the drawing

Step-by-step explanation:

The open circle means that it is not equal to and the arrow is going to the left because it is less than 1.

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lol that just means free points for me. ;)

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Step-by-step explanation:

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3 0
3 years ago
A poll of 793 adults aged 18 or older asked about purchases that they intended to make for the upcoming holiday season. One of t
Ira Lisetskai [31]

Answer:

(0.5793, 0.6363)

Step-by-step explanation:

Given that a  poll of 793 adults aged 18 or older asked about purchases that they intended to make for the upcoming holiday season. One of the questions asked about what kind of gift they intended to buy for the person on whom they would spend the most. Clothing was the first choice of 482 people.

i.e. sample proportion p= \frac{482}{793} =0.608

q=1-p = 0.392

Std error of p = \sqrt{pq/n} =0.0173

For 90% Z crtiical = 1.645

Hence margin of error = 1.645 * std error = 0.0285

Confidence interval 90%

= proportion ±margin of error

=(0.5793, 0.6363)

3 0
3 years ago
A padlock has a four-digit code that includes digits from 0 to 9, inclusive. What is the probability that the code does not cons
Eduardwww [97]
Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.

4 0
3 years ago
Read 2 more answers
Explain 15 is not a prime factor
MissTica

<em>1</em><em>5</em><em> </em><em>is</em><em> </em><em>not</em><em> </em><em>a</em><em> </em><em>prime</em><em> </em><em>factor</em><em> </em><em>as</em><em> </em><em>it</em><em> </em><em>is</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>1</em><em>,</em><em>3</em><em>,</em><em>5</em><em> </em><em>and</em><em> </em><em>1</em><em>5</em><em>.</em>

<em>Additional</em><em> </em><em>information</em><em>:</em>

<em>Prime</em><em> </em><em>numbers</em><em> </em><em>are</em><em> </em><em>those</em><em> </em><em>numbers</em><em> </em><em>which</em><em> </em><em>can</em><em> </em><em>only</em><em> </em><em>be</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>itself</em><em> </em><em>.</em><em>for</em><em> </em><em>instance</em><em>:</em><em>1</em><em>,</em><em>2</em><em>,</em><em>3</em><em>,</em><em>5</em><em>,</em><em>7</em><em> </em><em>,</em><em>1</em><em>1</em><em>,</em><em>1</em><em>3</em><em> </em><em>etc</em><em>.</em>

<em>Composite</em><em> </em><em>number</em><em> </em><em>s</em><em> </em><em>are</em><em> </em><em>those</em><em> </em><em>numbers</em><em> </em><em>which</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>also</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>other</em><em> </em><em>numbers</em><em>,</em><em>For</em><em> </em><em>instance</em><em>:</em><em> </em><em>4</em><em>,</em><em>6</em><em>,</em><em>8</em><em>,</em><em>1</em><em>0</em><em> </em><em>etc</em>

<em>Hope</em><em> </em><em>it </em><em>helps</em><em>.</em><em>.</em><em>.</em><em>.</em>

7 0
3 years ago
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