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Yanka [14]
3 years ago
14

What is compatible number to find two estimates that the quotient is between 1624÷3

Mathematics
1 answer:
Vladimir [108]3 years ago
8 0
1500÷3 =500


1800÷3 =600
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Pedro and Carmela are cleaning their kitchen. It takes Pedro 3 hours to clean the kitchen by himself, while Carmela can clean th
Igoryamba

Answer:

B

Step-by-step explanation:

Here, we want to choose the equation representing the time it will take if they worked together

Let the amount of tasks to be done be t

pedro will work at a rate of t/3

Camela will take t/2

So the joint rate will be

t/2 + t/3

This joint rate will be equal to 1

Thus, we have

t/2 + t/3 = 1

5 0
3 years ago
Find the equation that Passes through the point (2,3) and is parallel to the line y=5x-9. First find the equation of the line in
KATRIN_1 [288]
Since It's parallel to <span>y=5x-9 the slope of this new line is 5
so the equation looks like </span><span>y=5x+b
to find be we just replace x by 2 and y by 3
3=10-b so b = -7
the equation is </span><span>y=5x-7</span>
7 0
4 years ago
The probability that a randomly selected 2 2​-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0
Mnenie [13.5K]

Answer:

a. Probability = 0.97735

b. Probability = 0.92294

c. P(At\ Least\ One) = 1

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

P(Live) = 0.98861

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)\ and\ P(Live)

Probability = 0.98861 * 0.98861

Probability = 0.9773497321

Probability = 0.97735 <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)^n

Where n = 7

Probability = 0.98861^7

Probability = 0.92294324145

Probability = 0.92294 <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

P(At\ Least\ One) = 1 - P(None).

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

P(Not\ Live) = 0.01139

The probability that none of the 7 lives up to 3 is:

P(None) = P(Not\ Live)^7

P(None) = 0.01139^7

Substitute this value for P(None) in

P(At\ Least\ One) = 1 - P(None).

P(At\ Least\ One) = 1 - 0.01139^7

P(At\ Least\ One) = 0.99999999999997513055642436060443621

P(At\ Least\ One) = 1 ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0
3 years ago
Sales prices of baseball cards from the 1960s are known topossess a skewed-right distribution with a mean sale price of $5.25and
Serga [27]

Answer:

c. Skewed-right with a mean of $5.25 and a standard error of$0.28

Step-by-step explanation:

As we have given that Sales prices of baseball cards have a right-skewed distribution with a mean $5.25 and a standard deviation is $2.80.

Now, We know that if standard deviation = σ

then, standard error = \frac{\sigma}{\sqrt{n}}

As, we have standard deviation = $2.80

then standard error will be \frac{2.80}{\sqrt{100}}

⇒ Standard error = $0.28

Hence, Option (c) is the correct option.

6 0
4 years ago
Match each point label on the box it to its description.
Ahat [919]
A is the minimum value
B is the first quartile
The point with the line through it (next to C) is the median. 
The point to the outside of the box is the third quartile.
D is the maximum value.

Hope this helped.
4 0
3 years ago
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