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mylen [45]
3 years ago
9

What is the least common multiple of15&25

Mathematics
1 answer:
nikklg [1K]3 years ago
3 0
The LCM of 15 and 25 is 75.
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Write an equation for the function graphed above.
Rus_ich [418]

Where is the graph ??

Please provide the graph so that we can answer your questions ....

8 0
2 years ago
Susan was given the following number to convert to scientific notation: 12,050,000 Her answer was 1.25 x 10^7 , but it was marke
KATRIN_1 [288]

Answer:

C. She left out a zero between the two and the five

Step-by-step explanation:

Given that Susan is expected to convert 12,050,000 to scientific notation, what Susan should have done is shown below:

Count the number of decimal places from your right up to before the first digit, and place the decimal point after the first digit. The number of decimal places counted would determine the exponent we are to use.

Here, we have 7 decimal places, counting from the right to the left, therefore, the exponent would be a positive 7.

What Susan should have is:

1.205 × 10⁷ NOT 1.25 × 10⁷.

1.25 × 10⁷ = 12,500,000, which is different from 12,050,000.

Her paper was marked wrong because: "she left out a zero between the two and the five"

7 0
2 years ago
if a pitcher is full of lemonade. Which unit of liquid volume best describes the amount of lemonade in the pitcher? explain
sergij07 [2.7K]
You wouldn't use, cm,m,km,kg,or anything like that.. you would use capacity.. (fancy word for liquid) So, for liquid you would use L, litres or gallons, or millilitres.
7 0
3 years ago
Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0
3 years ago
$54.25+55 cents equalssssssssssssssssss
never [62]
$54.80 is the correct answer
3 0
3 years ago
Read 2 more answers
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