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3 years ago

How to find the angle?

Mathematics

1 answer:

kumpel [21]3 years ago

4 0

If everything on inside of a line is 180->
(Line) --------/---------- everything on top equals 180 and there should be a number on either side. For instance 180-50 for the top would be 130, the wide angle's angle. I answer everything I like and come across so sorry I am four weeks late.

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Four batsmen on a cricket team had a mean of 42.5 runs the other 7 batsmen had a mea of 18 what was the mean number of scored by

Ira Lisetskai [31]

Answer:

The mean score of the entire team is 26.909.

Step-by-step explanation:

Given that four batsmen had a mean of 42.5.

Let $\textbf{B}_{\textbf{i}$ , where $i = 1, 2, 3, \hdots, 11$ denote the batsmen.

From the first statement, we have:

$\frac{B_1 + B_2 + B_3 + B_4}{4} = 42.5$

$\implies B_1 + B_2 + B_3 + B_4 = 42.5 \times 4 = \textbf{170}$

Therefore, the sum of the scores of the first four batsmen is 170.

Now, from the second statement we have:

$\frac{B_5 + B_6 + B_7 + B_8 + B_9 + B_{10} + B_{11}}{7} = 18$

$\implies B_5 + B_6 + B_7 + B_8 + B_9 + B_{10} + B_{11} = 18 \times 7 = \textbf{126}$

That is, the sum of the scores of the remaining 7 batsmen is 126.

Now, to calculate the average(mean) of the entire team, we add the individual scores and divide it by 11.

$\implies \frac{B_1 + B_2 + B_3 + B_4 + B_5 + B_6 + B_7 + B_8 + B_9 + B_{10} + B_{11}}{11} = \frac{170 + 126}{11}$

$= \frac{296}{11}$

$= \textbf{26.90}$ which is the required answer.

8 0

3 years ago

Four polynomials are shown below:

Sloan [31]

I think it is NO. C. I holp it is help.

3 0

4 years ago

1. The teacher assigns homework after 7/10

asambeis [7]

Answer:

B) Likely

Step-by-step explanation:

Unlikely = < 50%

Likely = > 50%

Certain = 100%

Equally likely = 50%

8 0

3 years ago

Which expression is equivalent to StartFraction (3 m Superscript negative 1 Baseline n squared) Superscript negative 4 Baseline

koban [17]

Answer:

$3m^{10}n^{-11}$

Step-by-step explanation:

Given the expression $\frac{(3m^{-1}n^{2})^{-4} }{(2m^{-2}n)^{3} }$ , we will use laws of indices to get the equivalent expression as shown below;

According to one of the law of indices,

$\frac{a^{m} }{a^{n} } = a^{m-n} \ and\ (a^{m})^{n} = a^{mn}$

$\frac{(3m^{-1}n^{2})^{-4} }{(2m^{-2}n)^{3} }\\= \frac{3m^{4}n^{-8} }{2m^{-6}n^{3} }\\= 3m^{(4-(-6))} * n^{-8-3}\\ = 3m^{10}n^{-11}$

This gives the required expression

8 0

3 years ago

Read 2 more answers

Please solve the whole page please please

Leya [2.2K]

1d) ↑ (Do the same but turned if that's easier)

1a) 199

1b) 193

1c) 217

1e) Based on the box plot, we can for example, find the lowest and greatest values, the range (largest - smallest), the mean (all values / # values), and the median middle value or q2 given by the second line in the box.

2b) 7

2c) 9.5

2d) ↑

2e) because the data is not ordered, it is difficult to determine the quartile data. Once it is ordered like 6, 7, 7, 7, 8, 8, 9, 10, 10. The data can be visualized alot easier. With the box plot we can determine q3 of the data by looking at the 3rd line, using the whiskers of the box to find the minimum and maximum values.

You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.

In this case the ordered data would be: 4,6,8,8,9,11,12,14,14,16.

[4, 6, 8, 8, 9] [] [11, 12, 14, 14, 16]

4 is the minimum, 16 is the maximum, q1 is 8, q3 is 14, and median is 10.

4) ↑

You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.

In this case the ordered data would be:

45, 45, 50, 55, 60, 70, 70, 75, 80, 85

[45, 45, 50, 55, 60] [] [70, 70, 75, 80, 85]

45 is the minimum, 85 is the maximum, q1 is 50, q3 is 75, and median is 65.

6) ↑

4 0

2 years ago