I'm confused, this says the triangle shown below has an area of 50 units2 find the missing side. One side is the length witch is
1 answer:
Answer:
The height of the given triangle is 9.091 units.
Step-by-step explanation:
Area of a triangle = x base x height
From the given question,
area of the triangle = 50
length = 11 units
width = 11 units
So that,
50 = x 11 x height
100 = 11 x height
height =
= 9.091
The height of the given triangle is 9.091 units.
You might be interested in
Answer:
See Explanation
Step-by-step explanation:
(Please Find Diagram in the attachment)⇒Answer Drawing is Given There.
According to the question,
- Given that, The city of Plainview is building a new sports complex. The complex includes eight baseball fields, four soccer fields, and three buildings that have concessions and restrooms.
- Now, Arrange the structures in the sports complex using translations, reflections, and rotations so that the final arrangement satisfies each of these criteria:
- All the fields and buildings fit on the provided lot.
- Each field is adjacent to at least one building for ease of access.
- Two or more fields can be adjacent, but no two fields should share the same boundary (e.g., a sideline or a fence.)
- For safety reasons, no baseball field should have an outfield (the curved edge) pointed at the side (the straight edges) of another baseball field
Answer:
Step-by-step explanation:
The mean SAT score is , we are going to call it \mu since it's the "true" mean
The standard deviation (we are going to call it ) is
Next they draw a random sample of n=70 students, and they got a mean score (denoted by ) of
The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.
- So the Null Hypothesis
- The alternative would be then the opposite
The test statistic for this type of test takes the form
and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.
With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.