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2 years ago

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can somebody please answer these, my last post didn’t show the questions lol sorry but i really need help. please help and show

a little work. any question

1 answer:

soldi70 [24.7K]2 years ago

3 0

Answer:

hi answer is 666666 I am very sad:(

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Plz help asap, hehdjjdnejdjfj

Dimas [21]

Answer:

135

Step-by-step explanation:

angle 1 and angle 3 are have the same messure so angle 3 equals to 135

8 0

3 years ago

Hey, guys help me answer!

Sliva [168]

Hello and Good Morning/Afternoon

<u>Let's take this problem step by step.</u>

$5=\frac{1}{3}(5x-15)\\ 5=\frac{5}{3}x-5\\ 5+5= \frac{5}{3}x-5 +5\\10=\frac{5}{3}x\\ 10 * 3 = 3* \frac{5}{3}x\\30=5x\\\frac{30}{5} =\frac{5}{5}x\\ 6=x\\x=6$

<u>What did we do in this problem</u>

we ensured that both sides remained equally balanced in value
took everything step-by-step to make sure that there were no mistakes
to check it, we can plug it back into the original equation to see if it works

$5=\frac{1}{3}(5(6)-15)\\ 5=\frac{1}{3} (30-15)\\5=\frac{1}{3}*15\\ 5=5 \checkmark\\$

<u>Answer: x = 6</u>

Hope that helps!

#LearnwithBrainly

5 0

1 year ago

Read 2 more answers

The frequency distributions of two data sets are shown in the dot plots below.

fredd [130]

Answer:

1st, 4th and 6th statements are true.

Step-by-step explanation:

We have been given two dot-plots and we are asked to select all the true statements about given dot plots.

Let us write all the data points of both dot plots.

Data set 1:

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 7.

Data set 2:

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4.

1. The mean of data set 1 is greater than the mean of data set 2.

Let us find mean of both data sets.

$\text{Mean of data set 1}=\frac{1+1+1+1+1+2+2+2+2+2+2+2+2+3+3+3+4+4+7}{19}$

$\text{Mean of data set 1}=\frac{45}{19}$

$\text{Mean of data set 1}=2.3684\approx 2.37$

$\text{Mean of data set 2}=\frac{1+1+1+1+1+2+2+2+2+2+2+2+2+3+3+3+4+4}{18}$

$\text{Mean of data set 2}=\frac{38}{18}$

$\text{Mean of data set 2}=2.11$

We can see that mean of data set 1 is greater than mean of data set 2, therefore, 1st statement is true.

2. The mean of data set 1 is equal to the mean of data set 2.

Since mean of data set 1 is greater than mean of data set 2, therefore, 2nd statement is false.

3. The median of data set 1 is greater than the median of data set 2.

Since data set 1 has 19 data points, so median of data set 1 will be value of 10th data set, that is 2.

Our data set 2 has 18 data points, so median of data set 2 will be the average of 9th and 10th data points.

$\text{Median of data set 2}=\frac{2+2}{2}$

$\text{Median of data set 2}=\frac{4}{2}=2$

Since median of data set 2 is also 2, therefore, 3rd statement is not true.

4. The median of data set 1 is equal to the median of data set 2.

Since both data set's median is 2, therefore, 4th statement is true.

5. The standard deviation of data set 1 is smaller than the standard deviation of data set 2.

Using online SD calculator we get SD of data set 1 is 1.422 and SD of data set 2 is 0.936. Since 1.422 is greater than 0.936, therefore, 5th statement is false.

6. The data sets have the same interquartile range.

$Q_1\text{ of data set 1}=1$

$Q_3\text{ of data set 1}=3$

$\text{IQR of data set 1}=3-1=2$

$Q_1\text{ of data set 2}=1$

$Q_3\text{ of data set 2}=3$

$\text{IQR of data set 2}=3-1=2$

Since both data set's IQR is 2, therefore, 6th statement is true.

3 0

3 years ago

Read 2 more answers

How do you do this haaaa

Ratling [72]

Answer:

7000

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

In a race in which eleven automobiles are entered and there are no ties, in how many ways can the first three finishers come in?

viva [34]

Answer:

990 ways

Step-by-step explanation:

The total number of automobiles we have is 11.

Now, what this means is that for the first position , we shall be selecting 1 out of 11 automobiles, this can be done in 11 ways( 11C1 = 11!/(11-1)!1! = 11!/10!1! = 11 ways)

For the second position, since we have the first position already, the number of ways we can select the second position is selecting 1 out of available 10 and that can be done in 10 ways(10C1 ways = 10!9!1! = 10 ways)

For the third position, we have 9 automobiles and we want to select 1, this can be done in 9 ways(9C1 ways = 9!/8!1! = 9 ways)

Thus, the total number of ways the first three finishers come in = 11 * 10 * 9 = 990 ways

8 0

3 years ago