What expression can be used for estimating 868 divided by 28

Social studies Geometry P.11 Similarity and altitudes in right triangles CEZ You have priz If SU = 28 and UV = 6, what is TU? T

Alex777 [14]

Answer:

TU ≈ 12.96

Step-by-step explanation:

Using the Altitude on Hypotenuse theorem

(leg of outer triangle)² = (part of hypotenuse below it) × (whole hypotenuse)

TU² = UV × SU = 6 × 28 = 168 ( take square root of both sides )

TU = $\sqrt{168}$ ≈ 12.96 ( to the nearest hundredth )

7 0

2 years ago

Please help me with this homework

fiasKO [112]

Answer:

It will be $2.40

4 0

2 years ago

Help me please:<br> 3/4 - 7/9 + 2/3 =

mojhsa [17]

Answer:

3/4-7/9+2/3=0.63

Step-by-step explanation:

.........

8 0

3 years ago

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The Ferrari Testarossa and the Porsche 911 Turbo are 400 miles apart and headed straight toward each other. The Ferrari is going

ryzh [129]

Answer:

1.6 hours

Step by step explanation;

1 hour would make them only 150 miles apart, 60% of 120 and 60% of 130 added together is 150 mph, so 1.6 hours

Hope this helps :)

5 0

2 years ago

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

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