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CaHeK987 [17]
2 years ago
10

Which expressions are equivalent to the one below? Check all that apply. log7 1• log5 25

Mathematics
1 answer:
ELEN [110]2 years ago
3 0

Answer:

D

Step-by-step explanation:

Use logarithms properties:

\log_a1=0\\ \\\log_aa=1\\ \\\log_ab^k=k\log_ab

Now,

\log_71=0\\ \\\log_525=\log_55^2=2\log_55=2\cdot 1=2

and

\log_71\cdot \log_525=0\cdot 2=0

Hence, correct option is option D.

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30,45,x,100; The median is 51
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To solve for x, you must first understand how the median was calculated out of the given set of numbers. Without looking at the given median value, we can see that we cannot get the median by process of elimination since there are an even amount of numbers in this particular set. Therefore, we must average the two closest values to what should be the median.

In this case, the values are "45" and "x". If we pretend that we know the value of the variable "x" (for example we will pretend that x is 55), then we should have an equation that looks like this: (45+55) ÷ 2 = [median]. What this equation is doing is adding the two closest values to the median (45 and 55) and dividing it by 2, the number of values we are averaging. Now we can solve this equation and simplify it to 100 ÷ 2 which is 50, our median.

So if they give us the median instead of the x value, then we can rewrite the equation to fit your request: (45+x) ÷ 2 = 51. Now we can solve for x:

1. Multiply by 2
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2. Subtract 45
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The x value for your question is 57.
3 0
2 years ago
Read 2 more answers
A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556
MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

The formula to find the standard deviation:-

\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}

The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.

Then,  \overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560

Now, \sum_{i=1}^{10}(x_i-\overline{x})^2=0^2+2^2+(-4)^2+(-2)^2+0^2+(-4)^2+(-1)^2+1^2+5^2+3^2\\\\\Rightarrow\ \sum_{i=1}^{10}(x_i-\overline{x})^2=76

Then, \sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76

Hence, the  standard deviation of his measurements = 2.76 g

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2 years ago
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Answer:

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