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NARA [144]
3 years ago
12

As part of a screening process, computer chips must be operated in an oven at 145 °C. The starting temperature is 25 °C. After 1

5 minutes, the temperature is 85 °C. How many minutes will it take the oven to reach its specified operating temperature?
Mathematics
1 answer:
DaniilM [7]3 years ago
4 0

Answer:

30 minutes will it take to reach its specified operating temperature from 25°C.

Step-by-step explanation:

Initial Temperature of oven = 25°C

Time taken by oven to reach 85°C temperature = 15 minutes

We have to find time taken by oven to reach at temperature of 145°C

Increase in temperature in 15 minute = 85 - 25 = 60°C

So, Time take by oven to increase 60°C temperature = 15 min

Time take by oven to increase 1°C temperature = \frac{15}{60}=\frac{1}{4}\:minutes

Temperature we have to increase = 145 - 25 = 120°C

Time take by oven to increase 120°C temperature = \frac{1}{4}\times120=30\:minutes

Therefore, 30 minutes will it take to reach its specified operating temperature from 25°C.

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For f(x) = x + 7, find f(x) when x = 3 and when x = -5.​
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3 0
3 years ago
Consider the hypotheses below. Upper H 0​: mu equals 50 Upper H 1​: mu not equals 50 Given that x overbar equals 60​, s equals 1
Neko [114]

Answer:

We conclude that  \mu\neq 50.

Step-by-step explanation:

We are given that x bar = 60​, s = 12​, n = 30​, and alpha = 0.10

Also, Null Hypothesis, H_0 : \mu = 50

Alternate Hypothesis, H_1 : \mu \neq 50

The test statistics we will use here is;

                    T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~  t_n_-_1

where, X bar = sample mean = 60

              s = sample standard deviation = 12

              n = sample size = 30

So, Test statistics = \frac{60-50}{\frac{12}{\sqrt{30} } } ~ t_2_9

                             = 4.564

At 10% significance level, t table gives critical value of 1.311 at 29 degree of freedom. Since our test statistics is more than the critical value as 4.564 > 1.311 so we have sufficient evidence to reject null hypothesis as our test statistics will fall in the rejection region.

P-value is given by, P(t_2_9 > 4.564) = less than 0.05% {using t table}

Here also, P-value is less than the significance level as 0.05% < 10% , so we will reject null hypothesis.

Therefore, we conclude that \mu\neq 50.

7 0
3 years ago
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