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alexandr402 [8]

2 years ago

14

Using the table below, determine the unit rate in beats per minute.

1 answer:

Zepler [3.9K]2 years ago

7 0

Answer: Where is the table?

Step-by-step explanation: i wish i could help but im afraid there isnt a table

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The sum of two numbers is 5.036. One number has a4 in the tenths place and a 7in the thousandths place. The other number has a 1

wlad13 [49]

The sum of two numbers 5.036.
first number: one number has 4 in the tenths place, 7 in the thousandths place.
=> x.4x7
2nd number: 1 in ones place and 2 in the hundredths place.
=> 1.x2x
Now, we have the 2 numbers with given clues:
=> x.4x7 + 1.x2x, where x is the missing number.
Note that in adding numbers, we start from the right most.
=>x.4x7 + 1.x2x = 5.036
=> x.4x7 + 1.x29
=> x.407 + 1.x29
=> x.407 + 1.629
=> 3.407 + 1.629
=> 5.036

8 0

3 years ago

Bro what does this mean

dimulka [17.4K]

Answer:

A

Step-by-step explanation:

hope this helps

8 0

3 years ago

Read 2 more answers

The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p

VLD [36.1K]

Answer:

$3.13$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

$\chi^2_{1-\frac{\alpha}{2}}= 51.770$

Chi square critical value for upper tail =

$\chi^2_{\frac{\alpha}{2}}= 81.085$

80% confidence interval:

$\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}$

Putting values, we get,

$=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13$

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0

3 years ago

What is the slope of a line that is perpendicular to the graph of<br> y= 5x + 5?

AVprozaik [17]

-5/1 is the slope so the answer is -5 because you pick 2 points and use point slope formula or
y1-y2/x1-x2

3 0

3 years ago

Helpppppppppppppppppppp

Sloan [31]

Answer:

Sum of $\frac{7x}{x^2-4}$ and $\frac{2}{x+2}$ is $\mathbf{\frac{9x-4}{x^2-4}}$

Option B is correct answer.

Step-by-step explanation:

We need to find sum of $\frac{7x}{x^2-4}$ and $\frac{2}{x+2}$

Finding sum of $\frac{7}{x^2-4}$ and $\frac{2}{x+2}$ :

$\frac{7x}{x^2-4}+\frac{2}{x+2}$

We know that $x^2-4 =(x+2)(x-2)$

Replacing x^2-4

$\frac{7x}{(x+2)(x-2)}+\frac{2}{x+2}$

Now, taking LCM of (x+2)(x-2) and (x+2) we get (x+2)(x-2)

$=\frac{7x+2(x-2)}{(x+2)(x-2)}\\=\frac{7x+2x-4}{(x+2)(x-2)}\\=\frac{9x-4}{(x+2)(x-2)}\\=\frac{9x-4}{x^2-4}$

So, Sum of $\frac{7x}{x^2-4}$ and $\frac{2}{x+2}$ is $\mathbf{\frac{9x-4}{x^2-4}}$

Option B is correct answer.

6 0

2 years ago