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Delicious77 [7]
3 years ago
10

10 oranges and 20 grapefruits weigh 35 pounds total. 16 oranges and 10

Mathematics
1 answer:
lana [24]3 years ago
7 0

Answer:

B

Step-by-step explanation:

oranges weigh 0.5 and grapefruits weigh 1.5

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John must score 80% correct on a test to qualify for a job. His score was 36/40. Does he qualify? What is his percent correct?
Ulleksa [173]

Answer:

he qualifies for the job

4 0
3 years ago
What is the absolute value of 300-48t
AleksandrR [38]
The absoulte vaue of 300-48t is 30=29=-4
7 0
3 years ago
What is the volume of a cone that has a height of 7 mm and a radius of 3 mm? Use 3.14 to approximate pi and round the answer to
Anna [14]
Volume=1/3 times area of base (circle) times height

circle=3.14 times radius^2
circle=3^2 times 3.14
circle=9 times 3.14=28.26
times height (7)
197.82
times 1/3=65.94
answer is 65.94 mm^3
3 0
3 years ago
The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed.
Anton [14]

Answer:

a) The evidence suggest the true mean life of a biomedical device > 5500

b) (5487.94, +∞)

c) There is evidence to support that the mean is equal to 5500

Step-by-step explanation:

Here we have

(a) To test the hypothesis we have the claim that the mean life of biomedical devices > 5500

Therefore, we put the null Hypothesis which is the proposition that a difference does not exist. That is

H₀:  μ = 5500

Therefore, the alternative hypothesis will be

Hₐ:  μ > 5500

The test statistic is then found by;

t=\frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}} =\frac{5617.8-5500 }{\frac{234.5 }{\sqrt{15}}} \approx 1.95

The P value from the T table at df = n - 1 = 15 - 1 = 14 is

0.025 < P < 0.05

The P value is given as 0.036 from the T distribution at 14 derees of freedom df

Therefore, since P < α or 0.05, we reject the null hypothesis. That is we fail to reject Hₐ:  μ > 5500. The evidence suggest the true mean life of a biomedical device > 5500

(b) The 95% confidence interval is given as

CI=\bar{x}\pm t_\alpha \frac{s}{\sqrt{n}}

Which gives    t_\alpha =  \pm2.145

CI=5617.8\pm 2.145 \times \frac{234.5}{\sqrt{15}}

The confidence interval of the lower bound on the mean is then

(5487.94, +∞)

c) From the above result, we find that the mean of 5500 is contained in the range of the lower confidence on the mean. We can therefore, accept the null hypothesis. That is there is evidence to support that the mean is equal to 5500.

6 0
3 years ago
Find the 14th term of the geometric sequence 10,40,160
Galina-37 [17]

Answer:

671,088,640.

Step-by-step explanation:

To determine the 14th term of the geometric sequence that begins with 10, 40 and 160, knowing that each number is multiplied by 4, the following calculation must be performed, taking into account that the initial number (10) is multiplied by 4, and that said number must be potentiated 13 times to obtain the 14th term:

10 x (4 ^ 13) = X

10 x 67,108,864 = X

671,088,640 = X

Therefore, the 14th term of the geometric sequence will be 671,088,640.

3 0
3 years ago
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