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Mashutka [201]
2 years ago
14

A ladybugs length measures 2cm express this measurement in meters explain your thinking include an equation with an exponent in

your explanation
Mathematics
1 answer:
Ksivusya [100]2 years ago
7 0
All we really need to do here is to convert 2 cm to meters.  There are 100 cm in 1 meter, so the appropriate cm to meters conversion factor is 

1 meter
-----------
100 cm

Thus, 

2 cm       1 m
-------- * ----------- = (1/50) m = 0.02 m      or  2.0*10^(-2) m     (answer)
    1        100 cm
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Tony is pouring milk for student lunches. She knows that 3 quarts of milk contain 12 cups. How many cups can she pour with 7 qua
Evgesh-ka [11]

Answer:

7 quarts = 28 cups

10 cups = 2.5 quarts

Step-by-step explanation:

In order to solve the two different questions in the problem, starting with a ratio of the number of quarts to the number of cups would be helpful:

\frac{quarts}{cups}=\frac{3}{12}=\frac{1}{4}

So, for every 1 quart of milk, she can pour 4 cups.  Using this ratio, we can set up equivalent ratios, or proportions, to solve for the missing value.  For the first question, we need to know that amount of cups in 7 quarts:

\frac{1}{4}=\frac{7}{x}

We can solve for 'x' by multiplying the numerator and denominator by the same factor, in this case '7'.  So, 4 x 7 = 28 cups.

Next, we need to solve for the amount of quarts needed for 10 cups:

\frac{1}{4}=\frac{x}{10}

4 x 2.5 = 10, so we need to multiply the numerator by 2.5: 1 x 2.5 = 2.5 quarts.

5 0
2 years ago
9. a. Maurice says that 1079 ÷ 62 is 16 with a remainder of 87. a. Without seeing his work, how can you tell Maurice divided inc
mamaluj [8]

Answer:

a.  We can tell that Maurice divided incorrectly because if the dividend is 62, you can't have a remainder of 87, because 62 goes into 87 once, so the answer would be 17 with a remainder of 25.  B. YOu can use that fact to find the correct quotient, because 87/62 is 1, with a remainder of 25, so 16+1=17, so the answer is 17 with a remainder of 25.   Please vote the brainliest.

Step-by-step explanation:

8 0
2 years ago
How to find each angle indicated
Natalija [7]
The answer is 80 because all triangles equal 180
8 0
2 years ago
Read 2 more answers
2x-8y=6<br> y=-7- x<br> what are the coordinates
goblinko [34]
2x-8(-7-x)=6
2x+56+8x=6
10x=-50
x=-5

y=-7-(-5)
y=-2

(-5,-2)
5 0
3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
2 years ago
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