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Kruka [31]
3 years ago
13

An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before

they failed is 19.0 hours. The distribution of the lives approximated a normal distribution. The standard deviation of the distribution was 1.2 hours. About 95.44 percent of the batteries failed between what two values?a. 8.9 and 18.9
b. 12.2 and 14.2
c. 14.1 and 22.1
d. 16.6 and 21.4
Mathematics
1 answer:
joja [24]3 years ago
8 0

Answer:

option (d) 16.6 and 21.4

Step-by-step explanation:

Data provided in the question:

The mean life for a particular use before they failed = 19.0 hours

The distribution of the lives approximated a normal distribution

The standard deviation of the distribution = 1.2 hours

To find:

The values between which 95.44 percent of the batteries failed

Now,

In Normal distribution, the approximately 95% ( ≈ 95.44% of all values ) falls within 2 standard deviations of the mean

Therefore,

Upper limit = Mean + 2 × standard deviation

⇒ Upper limit = 19.0 + 2 × 1.2 = 21.4

Lower  limit = Mean - 2 × standard deviation

⇒ Lower  limit = 19 - 2 × 1.2 = 16.6

Hence,

the answer is option (d) 16.6 and 21.4

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neonofarm [45]

Here's link to the answer:

tinyurl.com/wpazsebu

8 0
3 years ago
What multiplies to be 280 and adds to be negative 36
photoshop1234 [79]
The two numbers are -18+ 2√(11) and -18 -2√(11).

-18+ 2√(11)+ [-18 -2√(11)]= -36
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Hope this helps~


4 0
3 years ago
Can someone help me with question e please ? ​
MAXImum [283]

Answer:

Step-by-step explanation:

a) The domain is R- all real numbers.

b) The domain is R- all real numbers.

c) we need

x^{2}-4\neq  0\\x^{2} \neq 4\\x\neq 2\\and\\x\neq -2

So the domain is R\{-2, 2}

d)

we need

x^{2}+x-2\neq  0\\x\neq 1\\and\\x\neq -2

So the domain is R\{-2, 1}

e)

we need: x-1>0 or x> 1

So the domain is (1,+∞)

6 0
3 years ago
. If n(A)=2 n(B)=3 n(AՈB) =24, what is the value of n(AUB)?​
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Step-by-step explanation:

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8 0
3 years ago
a study studied the birth weights of 1,999 babies born in the United States. The mean weight was 3234 grams with a standard devi
Aleks [24]

Answer:

1899

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 3234

Standard deviation = 871

Percentage of newborns who weighed between 1492 grams and 4976 grams:

1492 = 3234 - 2*871

So 1492 is two standard deviations below the mean.

4976 = 3234 + 2*871

So 4976 is two standard deviations above the mean.

By the Empirical Rule, 95% of newborns weighed between 1492 grams and 4976 grams.

Out of 1999:

0.95*1999 = 1899

So the answer is 1899

8 0
3 years ago
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