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Zigmanuir [339]
2 years ago
9

Explain how you can use a model to find 6x17

Mathematics
2 answers:
777dan777 [17]2 years ago
7 0
Maybe by useing cubes like 6 groups of 17 or u can also use a number line like the elementary skills
TiliK225 [7]2 years ago
4 0

Answer:

The required model is shown below:

Step-by-step explanation:

Consider the provided information.

We need to use partial products model to solve 6×17

The partial product is the product of parts of each factor.

Here Multiplicand is 6 and Multiplier is 17

We can rewrite 17 as 1 ten and 7 ones. i.e 17=10+7

Therefore,

6×17 = 6(10+7)

6×17 = 6×10+6×7  [Distributive property]

6×17 = 60+42 = 102

Hence, the required model is shown below:

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Based on Pythagorean identities, which equation is true? sine squared theta minus 1 = cosine squared theta secant squared theta
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Answer:

Cot^2\theta-Csc^2\theta=-1

Step-by-step explanation:

The only Pythagorean identities are:

1. \ Sin^2\theta+Cos^2\theta=1

2. \ 1+Tan^2\theta=Sec^2\theta

3. \ 1+Cot^2\theta=Csc^2\theta

4. \ Sin^2\theta=1-Cos^2\theta\\  \ \ Cos^2\theta=1-Sin^2\theta

 

Therefore,Cot^2\theta-Csc^2\theta=-1 is correct as it's one of the pythagorean identities.

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2 years ago
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The track team is trying to reduce their time for a relay race. First they reduce their time by 2.1 minutes. Then they are able
konstantin123 [22]

Answer: 15.7 minutes

Step-by-step explanation:

Let x be the time in the beginning (in minutes).

Given: The track team is trying to reduce their time for a relay race.

First they reduce their time by t_1=2.1 minutes.

Then they are able to reduce that time by t_2=10

If their final time is 3.96 minutes, then

x-t_1-t_2=3.6\\\Rightarrow\ x=3.6+t_1+t_2\\\Rightarrow\ x=3.6+2.1+10\\\Rightarrow\ x= 15.7

Hence, their beginning time was 15.7 minutes.

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You are planning to buy one of two brands of sofas, which you hope to use over the next twenty years. brand j costs $975 and las
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Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random
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Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes P_B(A).

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
P(A|B)= \frac{P(A \cap B)}{P(B)}

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is \frac{1}{3}

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

</span><span>Thus P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}

Therefore, P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}


Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}

</span><span>P(D)= \frac{1}{3}</span><span>

Therefore, P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}</span>
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