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anzhelika [568]

3 years ago

7

Martin is making squares by arranging 26 sticks.How many squares can he make? Write and solve the number fact you used to find t

he answer, and explain your reasoning.

1 answer:

Greeley [361]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

eduardo has 4 1/2 yard of rope light. Exactly how many more yard does he need to finish the car's ceiling

Over [174]

Answer:

2 1/6

Step-by-step explanation:

Because the denominators are different, and there are whole numbers involved, then first make the fractions improper.

4 1/2 -> 9/2

6 2/3 -> 20/3

Next, make the denominators the same by multiplying each fraction by the denominator of the other.

9/2 * 3 = 27/6

20/3 * 2 = 40/6

Then you can subtract the numerators because the denominators are the same:

40/6-27/6= 13/6

Then you can convert to a mixed fraction if need be.

3 0

2 years ago

Which ordered pair is a solution of the equation y-4=7(x-6) A.(5,4) B.(6,5) C.(5,4)(6,5) D.Neither

Natasha_Volkova [10]

Answer:

D. Neither

Step-by-step explanation:

3 0

3 years ago

Two Social Security numbers (see Exercise 8.12) match zeros if a digit of one number is zero iff the corresponding digit of the

Roman55 [17]

Answer:

Proved

Step-by-step explanation:

From the given parameters, we have:

$n = 9$ i.e. the length of each security numbers

$r = 2$ i.e. 2 security numbers

Required

In 513 security numbers, 2 must have matching zeros

To do this, we make use of Pigeonhole principle.

First, we calculate the number of all security numbers not having matching zeros.

Each of the 9 digits can be selected in 2 ways.

2 ways implies that each digit is either 0 or not

So, total selection is:

$Total = 2^9$

$Total = 512$

Apply Pigeonhole principle

The principle states that: suppose there are n items in m containers, where $n>m$ , then there is at least one container that contains more than 1 item.

This means that if there are 512 security number without matching zeros, then there is 1 (i.e. 512 + 1) with matching zeros.

$512 + 1 = 513$

8 0

3 years ago

Teddy makes 32 fluid ounces of hot cocoa. He pours equal amounts of cocoa into 5 cups. The amount of hot cocoa in each cup will

alisha [4.7K]

B is the answer trust me lol

8 0

3 years ago

Somebody please assist me here

Anettt [7]

The base case of $n=1$ is trivially true, since

$\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)$

but I think the case of $n=2$ may be a bit more convincing in this role. We have by the inclusion/exclusion principle

$\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)$

with equality if $E_1\cap E_2=\emptyset$ .

Now assume the case of $n=k$ is true, that

$\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)$

We want to use this to prove the claim for $n=k+1$ , that

$\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)$

The I/EP tells us

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)$

and by the same argument as in the $n=2$ case, this leads to

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})$

By the induction hypothesis, we have an upper bound for the probability of the union of the $E_1$ through $E_k$ . The result follows.

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)$

5 0

2 years ago