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1 year ago

Jeremy had 3/4 of a submarine sandwich and gave

1 answer:

5 0

Actual question is "<span>Jeremy had 3/4 of a submarine sandwich and gave his friend 1/3 of it. What fraction of the sandwich did the friend receive?"

Solution:
1/3 of 3/4 of a submarine sandwich = 1/3x3/4 = 3/12 = 1/4
so jermy gave 1/4 of the submarine sandwich to his friend. </span>

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100 random samples were taken from a large population. A particular numerical characteristic of sampled items was measured. The

Marta_Voda [28]

Answer:

The median is $Median = 0.903$

Step-by-step explanation:

From the question we are told that

The sample size is n = 100

The $1^{st} \to 45^{th}$ measurements is $= 0.859 \to 0.900$

Generally since that after 0.900 we have 0.901 , then the

$46^{th} \ measurement \ is \ 0.901$

in the same manner the $47^{th} \ measurement \ is \ 0.902$ ,

Given that 0.902 was observed three times it means that

$47^{th},48^{th},49^{th} \ measurement \ is \ 0.902$ ,

Given that 0.903 was observed two times it means that

$50^{th},51^{th} \ measurement \ is \ 0.903$ ,

Given that 0.903 was observed four times it means that

$52^{nd},53^{rd},54^{th},55^{th} \ measurement \ is \ 0.904$ ,

Given that the highest measurement is 0.958 then then the $56^{th} \to 100^{th} \ measurement \ is \ between \ 0.905 \to 0.958$

Generally the median is is mathematically represented as

$Median = \frac{ [\frac{n^{th}}{2}] + [(\frac{n}{2})^{th} + 1 ]}{2}$

=> $Median = \frac{ [\frac{100^{th}}{2}] + [(\frac{100}{2})^{th} + 1 ]}{2}$

=> $Median = \frac{ [50^{th}] + [51^{th} ]}{2}$

=> $Median = \frac{ 0.903 + 0.903}{2}$

=> $Median = 0.903$

8 0

2 years ago

Use integers that are that are closest to the number in the middle. < - v119

Leni [432]

Answer:

-11 and -10

Step-by-step explanation:

-11² = -121

-(√119)² = -119

-10² = -100

_____

-√119 is between -11 and -10

3 0

2 years ago

Jacob solves the system of equations by forming a matrix equation.

Lyrx [107]

Simultaneous equations can be solved using inverse matrix operation.

The complete steps of Jacob's solution are:

$\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$

We have:

$4x + y = 2$

$-2x + 4y = -22$

Calculate the determinant of $\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

$|A| = 4 \times 3 -1 \times -2$

$|A| = 12 +2$

$|A| = 14$

So, the inverse matrix becomes

$A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

Replace the first column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of x

$x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]$

So, we have:

$x = \frac{1}{14}(2 \times 3 - 1 \times -22)$

$x = \frac{1}{14}(6 +22)$

$x = \frac{1}{14}(28)$

$x = 2$

Replace the second column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of y

$y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right]$

So, we have:

$y = \frac{1}{14}(4 \times -22 - 2 \times -2)$

$y = \frac{1}{14}(-88 +4)$

$y = \frac{1}{14}(-84)$

$y = -6$

Hence, the complete process is:

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$

Read more about matrices at:

brainly.com/question/11367104

6 0

1 year ago

Abdullah is a quality control expert at a factory that paints car parts. He knows that 20\ , percent of parts have an error in t

-Dominant- [34]

The error rate has decreased after changing the painting process.

<u>Step-by-step explanation:</u>

Abdulla knows that 20 percent of the parts have an error in their painting. After suggesting changes in painting process, he wants to know whether the error rate has changed.

Number of parts in the random sample=400400400

Number of parts that had an error=606060

We have to determine what percentage of 400400400 is 606060

$606060=x/100 \times 400400400\\=0.15%$

After changing the painting process 0.15% of parts have error.

The previous percentage was 20.Hence the error rate has clearly changed.

3 0

1 year ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

1 year ago