HELP ME PLS !
1 answer:
Answer:
Step-by-step explanation:
1. The length of the fish a fisherman catches, f, must be at least 8 inches for him to keep it.
8 inches <_f
2. The number of days, d, of sunshine is not 32.
i don't get this one sorry
3. The value, v, of the car is less than $8,400.
v<$8,400
4. The number of people, p, that a café can seat at once is no more than 53.
p<_53 people
5. The time, t, a customer has left on a parking meter is at least 13 minutes.
t<_13 minutes
6. The bill, b, was less than $56.
b<_$56
17, and 19 might be wrong. (just which way the arrow goes)
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Answer:
<em>P= 5.5 cm</em>
Step-by-step explanation:
<u>The Perimeter of Plane Shapes</u>
If we are given a plane shape, the perimeter is the measurement of the boundary of it. In case the shape is made of lines, the perimeter is the sum of the lengths of all of them.
We are given an irregular shape, we only need to add all the sides. But we don't have all the individual sides. Let's write the expression for the perimeter.
P=AE+ED+DC+CB+BA
We have: AE + DC = 1 1/5 cm, AB = 1 3/4 cm, DE = 1 1/4 cm, and BC = 1 3/10 cm. Note we don't have AE or DC, but we have their sum, so
P=AE+DC+ED+CB+BA
Recall that AB=BA, the order of the letters is not important. The perimeter is
We add the integer parts separate from the fractions
Now we use the common denominator (Least Common Multiple or LCM ) which is 20:
Operating
Simplifying
Answer:
No, there is not enough evidence at the 0.02 level to support the manager's claim.
Step-by-step explanation:
We are given that the company's promotional literature states that 68% of the chips fail in the first 1000 hours of their use. Also, a sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use.
And the quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage, i.e;
Null Hypothesis, : p = 0.68 {means that the actual percentage that fail is same as the stated percentage}
Alternate Hypothesis, : p 0.68 {means that the actual percentage that fail is different from the stated percentage}
The test statistics we will use here is;
T.S. = ~ N(0,1)
where, p = actual percentage of chip fail = 0.68
= percentage of chip failed in a sample of 900 chips = 0.66
n = sample size = 900
So, Test statistics =
= -1.267
Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have sufficient evidence to accept null hypothesis.
Therefore, we conclude that the actual percentage that fail is same as the stated percentage and the manager's claim is not supported.