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almond37 [142]
2 years ago
14

Find the area of the figure

Mathematics
1 answer:
Vitek1552 [10]2 years ago
3 0

Answer:

30 square units

Step-by-step explanation:

split shape into other shapes you know how to find the areas of.

the top half is made of 2 triangles and a rectangle, and the bottom half is one large rectangle

You might be interested in
What is the value of 5(a + b) when a=1 and b=6 ​
lana [24]

Answer: 35

Step-by-step explanation:

We substitute a for 1 and b for 6.

5(1+6)=5(7)=35

5 0
3 years ago
A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found
joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample​ mean is found to be 113​ and the sample standard​ deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

                               P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = 113

             s = sample standard​ deviation = 10

             n = sample size = 13

             \mu = population mean

<em>Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.356 < t_1_2 < 1.356) = 0.80  {As the critical value of t at 12 degree

                                          of freedom are -1.356 & 1.356 with P = 10%}  

P(-1.356 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.356) = 0.80

P( -1.356 \times }{\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.356 \times }{\frac{s}{\sqrt{n} } } ) = 0.80

P( \bar X-1.356 \times }{\frac{s}{\sqrt{n} } } < \mu < \bar X+1.356 \times }{\frac{s}{\sqrt{n} } } ) = 0.80

<u>80% confidence interval for </u>\mu = [ \bar X-1.356 \times }{\frac{s}{\sqrt{n} } } , \bar X+1.356 \times }{\frac{s}{\sqrt{n} } }]

                                           = [ 113-1.356 \times }{\frac{10}{\sqrt{13} } } , 113+1.356 \times }{\frac{10}{\sqrt{13} } } ]

                                           = [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

<u>80% confidence interval for </u>\mu = [ \bar X-1.333 \times }{\frac{s}{\sqrt{n} } } , \bar X+1.333 \times }{\frac{s}{\sqrt{n} } }]

                                              = [ 113-1.333 \times }{\frac{10}{\sqrt{18} } } , 113+1.333 \times }{\frac{10}{\sqrt{18} } } ]

                                               = [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

<u>98% confidence interval for </u>\mu = [ \bar X-2.681 \times }{\frac{s}{\sqrt{n} } } , \bar X+2.681 \times }{\frac{s}{\sqrt{n} } }]

                                              = [ 113-2.681 \times }{\frac{10}{\sqrt{13} } } , 113+2.681 \times }{\frac{10}{\sqrt{13} } } ]

                                               = [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed because t test statistics is used only when the data follows normal distribution.

6 0
2 years ago
Can a sequence be both arithmetic and geometric?
artcher [175]
An aritmetic sequence is like this
a_n=a_1+d(n-1) where a1=first term and d=common difference

geometric is a_n=a_1(r)^{n-1} where a1=first term and r=common ratio


can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same

therfor we need to solve d(n-1)=(r)^{n-1}
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?

when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal

the answer is no, a sequence cannot be both aritmetic and geometric
4 0
3 years ago
What is the product of 2p + q and –3q – 6p + 1?
Umnica [9.8K]
-12p^2 - 12pq + 2p -3q^2 + q
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3 years ago
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3 years ago
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