What is the range of possible sizes for side x?
2 answers:
Check the picture below.
the idea being that sides of 2.5 andd 8 are fixed, what's are the possible lengths of "x"?
well, if we open the fixed sides, we'll end up with a 3rd side of 2.5 + 8 = 10.5, however if "x" is ever 10.5, the figure will become a flat-line and we'd be losing the triangle.
and if we close the fixed sizes, the longest "x" could be is 8 - 2.5 = 5.5, but the same, it can't be 5.5 exactly otherwise the figure will turn to a flat-line.
5.5 < x < 10.5.
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Answer:
The number of onion rings is <u>3</u> in the combination meal and the number of sliders is <u>2</u> in the combination meal.
Step-by-step explanation:
Given:
At a particular restaurant, each onion ring has 65 calories and each slider has 250 calories. A combination meal with onion rings and sliders has a total of 5 onion rings and sliders altogether and contains 695 calories.
Now, to find the number of onion rings in the combination meal and the number of sliders in the combination meal.
Let the number of onion rings in the combination meal be
And let the number of sliders in the combination meal be
<u><em>So, the total number of onion rings and sliders are:</em></u>
Now, the total number of calories each onion rings and sliders has:
Substituting the value of from equation (1) we get:
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<em>Subtracting both sides by 1250 we get:</em>
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<em>Dividing both sides by -185 we get:</em>
<u><em>The number of onion rings in the combination meal = 3.</em></u>
Now, substituting the value of in equation (1) we get:
<u><em>The number of sliders in the combination meal = 2.</em></u>
Hence, the number of onion rings is 3 in the combination meal and the number of sliders is 2 in the combination meal.
Answer:
The answer is below
Step-by-step explanation:
Twenty-five blood samples were selected by taking every seventh blood sample from racks holding 187 blood samples from the morning draw at a medical center. The white blood count (WBC) was measured using a Coulter Counter Model S. The mean WBC was 8.636 with a standard deviation of 3.9265. (a) Construct a 90% confidence interval for the true mean using the FPCF. (Round your answers to 4 decimal places.) The 90% confidence interval is from to
Answer:
Given:
Mean (μ) = 8.636, standard deviation (σ) = 3.9265, Confidence (C) = 90% = 0.9, sample size (n) = 25
α = 1 - C = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05
From the normal distribution table, The z score of α/2 (0.05) corresponds to the z score of 0.45 (0.5 - 0.05) which is 1.645
The margin of error (E) is given by:
The confidence interval = μ ± E = 8.636 ± 1.2918 = (7.3442, 9.9278)
The 90% confidence interval is from 7.3442 to 9.9278