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3 years ago

Find the volume: 7cm 16 cm 8 cm

Mathematics

2 answers:

USPshnik [31]3 years ago

8 0

Answer: Greetings my dude. The volume of the triangular prism is around 1,138 Centimeters Cubed.

Hope this helps!

alukav5142 [94]3 years ago

3 0

Answer:

volume = length * breadth * height

= 16*7*8

=896

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Cos 0 = 12/15. find sin 0.

valentina_108 [34]

Answer:

$sin \theta = \frac{3}{5}$

Step-by-step explanation:

$cos \theta = \frac{12}{15}\\\\cos^2 \theta = 1 - sin^2 \theta\\\\sin^2\theta = 1 - cos^2 \theta\\\\$

$= 1 - (\frac{12}{15})^2\\\\= 1 - \frac{144}{225}\\\\= \frac{225 - 144}{225}\\\\=\frac{81}{225}$

$sin \theta = \sqrt{\frac{81}{225}} = \frac{9}{15} = \frac{3}{5}$

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3 years ago

Given point (-6, -2) and a slope of 5, write an equation in slope-intercept form.

Finger [1]

Answer:

y=ax+b

y=5(-6)+28=-2

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4 years ago

Each year, taxpayers are able to contribute money to various charities via their IRS tax forms. The following list contains the

monitta

Answer:

A. median and mode

B. Mean

C. Mean

d. Positively skewed

Step-by-step explanation:

a. Mean = sum of all values / numbers of values = 1804 / 23 = 78.4. this eliminates the mean and leaves the median and mode, both of which are 62.

Median = middlemost value = 62

Mode = most frequent value = 62

b. Only the Mean would be affected. It would move from 78.4 to 80.6. (calculated as 1853 / 23). The median and mode would not be affected as the 62 would remain the middlemost value and the most frequent.

c. Removing the largest value would result in the mean changing. The mean would now be 73 (calculated as 1607 / 22). Since the values are no longer 23, but now 22, the median would be the average of the 2 middlemost values, ((62+62) / 2) which would still result in 62 as the median.

d. we use the Five number summary:

Median / Interquartile Range = 62

First Quartile / Q1 = (51 +57) / 2 = 54

Third Quartile / Q3 = (99 + 104) / 2 = 101.5

Minimum = 2

Maximum = 197 ; therefore Range = (197 - 2) = 195.

Please see attached box and whisker plot. The distribution is positively skewed.