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2 years ago

13

There are 928 possible words for a spelling bee. If Levi studies 40 words per day, how long will he take to study all of the wor

ds?

1 answer:

LUCKY_DIMON [66]2 years ago

4 0

The answer is 24 days for this question

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Identify the errors the student made

uranmaximum [27]

Answer:

Second step: 2/2 does not give you 2.

It would be 2x + 1 - 4 = 3x - 3

Since he mess up on the second step, the following steps are also messed up

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2 years ago

PLEASE HELP!!!!!!!!!!

erik [133]

Answer: D

Step-by-step explanation:

7 0

2 years ago

Before the distribution of certain statistical software, every fourth compact disk (CD) is testedfor accuracy. The testing proce

pishuonlain [190]

Answer:

P(T∩E) = 0.017

Step-by-step explanation:

Since every fourth CD is tested. Thus if T is the event that represents 4 disks being tested,

P(T) = 1/4 = 0.25

Let Fi represent event of failure rate. So from the question,

P(F1) = 0.01 ; P(F2) = 0.03 ; P(F3) =0.02 ; P(F4) = 0.01

Also Let F'i represent event of success rate. And we have;

P(F'1) = 1 - 0.01 = 0.99 ; P(F'2) = 1 - 0.03 = 0.97; P(F'3) = 1 - 0.02 = 0.98; P(F'4) = 1 - 0.01 = 0.99

Since all programs run independently, the probability that all programs will run successfully is;

P(All programs to run successfully) =

P(F'1) x P(F'2) x P(F'3) x P(F'4) =

0.99 x 0.97 x 0.98 x 0.97 = 0.932

Now, that all 4 programs failed will be = 1 - 0.932 = 0.068

Let E be denote that the CD fails the test. Thus P(E) = 0.068

Now, since testing and CD's defection are independent events, the probability that one CD was tested and failed will be =P(T∩E) = P(T) x P(E)= 0.25 x 0.068 = 0.017

8 0

3 years ago

The information below describes a data plan for a storage device.

Vesna [10]

Answer

A) C=5+2d

Step-by-step-explanation

3 0

3 years ago

A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute

Katyanochek1 [597]

Answer:

Step-by-step explanation:

Given that:

sample size n = 36

standard deviation = 10.1

level of significance ∝ = 0.10

The null hypothesis and the alternative hypothesis can be computed as follows:

$H_0 : \sigma = 10$

$H_1 : \sigma \neq 10$

The test statistics can be computed as follows:

$X^2 = \dfrac{(n -1)s^2 }{\sigma ^2}$

$X^2 = \dfrac{(36 -1)10.1^2 }{10^2}$

$X^2 = \dfrac{(35)102.01 }{100}$

$X^2 = \dfrac{3570.35 }{100}$

$X^2 =35.704$

degree of freedom = n - 1 = 36 - 1 = 35

Since this test is two tailed .

The P -value can be determined by using the EXCEL FUNCTION ( = 2 × CHIDIST(35.7035, 35)

P - value = 2 × 0.435163515

P - value = 0.8703 ( to four decimal places)

Decision Rule : To reject the null hypothesis if P - value is less than the 0.10

Conclusion: We fail to reject null hypothesis ( accept null hypothesis) since p-value is greater than 0.10 and we conclude that there is sufficient claim that the normal range of pulse rates of adults given as 60 to 100 beats per minute resulted to a standard deviation of 10 beats per minute.

3 0

3 years ago