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KengaRu [80]
2 years ago
10

Help me with this people

Mathematics
2 answers:
Mice21 [21]2 years ago
7 0

Answer: 235 cm^2

Explanation: you can cut the shape into two smaller ones and then solve for each smaller shapes area. to find out the length of the top piece with 11 as length you can subtract nine from fourteen. from their you calculate the area of the top shape (11x5=55) and then the bottom shape (9x20=180). from their just add the two areas to get 235.

grigory [225]2 years ago
7 0

Answer:

246 cm squared

Step-by-step explanation:

u hv to break up the big shape into little shapes so you would do 20x9 and 20-14 which is 6 and do 6x11 and then add them up

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Is the relationship between 7s in 7,742 ant the 7s in 7,785 different in anyway? Explain
melisa1 [442]
The relationship between the 7's is not different in any way because the sevens are all in the same place. The sevens are in the hundreds and thousands place.
4 0
3 years ago
It is 7 km from Kerry's house to the mall. About what is that distance in miles?
gavmur [86]

1 km = 0.62137... mile (rounded)

7 km = 4.3296... miles (rounded)

7 0
3 years ago
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A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for
tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of <em>t</em> are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29

s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11

s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}

  =\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}

  =-4.123

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}

   =\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}

   =26.55\\\approx 27

Compute the critical value for <em>α</em> = 0.05 as follows:

t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052

*Use a <em>t</em>-table for the values.

Thus, the critical values of <em>t</em> are ± 2.052.

3 0
2 years ago
A.Calculate the mean,median and mode.(3 points each) 1.)1,2,3,4,5 2.)2,3,4,5,6,6 3.)6,7,5,4,5,6,2,5
zlopas [31]

Answer:

Step-by-step explanation:

1.)1,2,3,4,5

mean=sum of all values/number of values

         =1+2+3+4+5/5

         =15/5

mean=3

Mode :

In the given data, no observation occurs more than once.

Hence the mode of the observations does not exist, means mode=0.

Median

1,2,3,4,5

Middle value is 3 so the median is 3.

2.)2,3,4,5,6,6

mean=sum of all values/number of values

         =2+3+4+5+6+6/6

         =26/6

mean =4.33

Mode

      is that value of the observation which occurs maximum number of times so here mode is 6.

Median

2,3,4,5,6,6

 4+5/2

9/2

median=4.5

3.)6,7,5,4,5,6,2,5

mean=sum of all values/number of values

         =6+7+5+4+5+6+2+5/8

        =40/8

mean =5

Mode

is that value of the observation which occurs maximum number of times so here mode is 5

Median

2,4,5,5,5,6,6,7

5+5/2

10/2

median=5

 

5 0
3 years ago
Estimate the product of 38*249
AnnZ [28]
The correct answer is 10,000
5 0
3 years ago
Read 2 more answers
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