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Lynna [10]
2 years ago
12

Which of the following is true when constructing a line graph?

Mathematics
1 answer:
Marat540 [252]2 years ago
3 0

Answer:

B

Step-by-step explanation:

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Determine whether the given procedure results in a binomial distribution​ (or a distribution that can be treated as​ binomial).
timofeeve [1]

Answer:

The procedure results in a binomial distribution

Step-by-step explanation:

If all 120 couples used the YSORT method, it is fair to assume that the probability of a baby being a boy or a girl are constant through all trials (120 couples). Assuming 120 randomly selected couples, there is a fixed number of independent trials. Finally, since the babies can only be a boy or a girl, the binary condition is satisfied and thus, the distribution is binomial

3 0
3 years ago
Factor the quadratic expression completely 15x^-4x-4
In-s [12.5K]

Answer:

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

Step-by-step explanation:

4 0
3 years ago
$10,000 at an annual rate of 7%, compounded semi-annually, for 2 years
forsale [732]

Answer:

\$13,107.96

Step-by-step explanation:

Since interest is compounded semi-annually (half a year or 6 months), in a spawn of 2 years, the interest will have been compounded 4 times. As given in the problem, each time the interest is compounded, the new balance will be 107% or 1.07 times the amount of the old balance.

Therefore, we can set up the following equation to find the new balance after 2 years:

\text{New balanace}=10,000\text{ (old balance)}\cdot 1.07\cdot 1.07\cdot 1.07\cdot 1.07,\\\text{New balanace}=10,000\cdot 1.07^4=\boxed{\$13,107.96}

8 0
3 years ago
What are the numbers from least to greatest based on the -7 ,4, -2, 3
Anastaziya [24]

Answer:

<em>It would be -7,-2,3,4.</em>

8 0
3 years ago
Read 2 more answers
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
3 years ago
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