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1 year ago

12

The table below represents the closing prices of stock ABC for the last five

1 answer:

Ilia_Sergeevich [38]1 year ago

7 0

Answer:

b

Step-by-step explanation:

just trust me

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A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s

amid [387]

Answer:

The correct conclusion is:

<em>"The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches."</em>

Step-by-step explanation:

A doctor is measuring the average height of male students at a large college.

The doctor measures the heights, in inches, of a sample of 40 male students from the baseball team.

Using this data, the doctor calculates the 95% confidence interval (63.5, 74.4).

The following conclusions is valid:

<em>"The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches."</em>

Since we know that the confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.

For the given case, the confidence level is 95% and the corresponding confidence interval is (63.5, 74.4) which represents the true mean of heights for male students at the college where the doctor measured heights.

Therefore, it is valid to conclude that the doctor is 95% confident that the mean height of male students at the college is within the interval of (63.5, 74.4).

6 0

2 years ago

Find the GCF of 44j5k4 and 121j2k6.

BartSMP [9]

Answer:

D. $11j^2k^4$

Step-by-step explanation:

We are asked to find the GCF of $44j^5k^4\text{ and }121j^2k^6$ .

Since we know that GCF of two numbers is the greatest number that is a factor of both of them.

First of all we will GCF of 44 and 121.

Factors of 44 are: 1, 2, 4, 11, 22, 44.

Factors of 121 are: 1, 11, 11, 121.

We can see that greatest common factor of 44 and 121 is 11.

Now let us find GCF of $j^5\text{ and }j^2$ .

Factors of $j^5$ are: $j*j*j*j*j$

Factors of $j^2$ are: $j*j$

We can see that greatest common factor of $j^5\text{ and }j^2$ is $j*j=j^2$ .

Now let us find GCF of $k^4\text{ and }k^6$ .

Factors of $k^4$ are: $k*k*k*k$

Factors of $k^6$ are: $k*k*k*k*k*k$

We can see that greatest common factor of $k^4\text{ and }k^6$ is $k*k*k*k=k^4$ .

Upon combining our all GCFs we will get,

$11j^2k^4$

Therefore, GCF of $44j^5k^4\text{ and }121j^2k^6$ is $11j^2k^4$ and option D is the correct choice.

3 0

2 years ago

Read 2 more answers

The velocity of an object in meters per second varies directly with time in seconds since the object was dropped, as represented

larisa86 [58]

Answer: 10.2

Step-by-step explanation:

3 0

2 years ago

Determine the positive real root of ln(x²) = 0.7

timama [110]

Answer:

i have no clue.

Step-by-step explanation:

i am very tired while doing a math link so please, don't give me any trouble with this :)

5 0

2 years ago

A jet flies 425 km from Ottawa to Québec at rate v + 60. On the return flight, the

Marina CMI [18]

Answer:

a. $\frac{- 42,500}{(v + 60)(v - 40)}$

Step-by-step Explanation:

Given:

Distance Ottawa to Québec = 425 km

Initial flight rate = v + 60

Return flight rate = v - 40

$t = \frac{d}{r}$

Required:

Flight times difference of the initial and return flights

Solution:

=>Flight time of the initial flight:

$t = \frac{d}{r}$

$t = \frac{425}{v + 60}$

=>Flight time of the return flight:

$t = \frac{425}{v - 40}$

=>Difference in flight times:

$\frac{425}{v + 60} - \frac{425}{v - 40}$

$\frac{425(v - 40) -425(v + 60)}{(v + 60)(v - 40)}$

$\frac{425(v) - 425(40) -425(v) -425(+60)}{(v + 60)(v - 40)}$

$\frac{425v - 17000 -425v - 25500}{(v + 60)(v - 40)}$

$\frac{425v - 425v - 17000 - 25500}{(v + 60)(v - 40)}$

$\frac{- 42,500}{(v + 60)(v - 40)}$

3 0

2 years ago