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2 years ago

Jude graphed the inequality s < -9,="" as="" shown.="">

Mathematics

1 answer:

pishuonlain [190]2 years ago

5 0

Answer:

No, Jude’s graph is incorrect. The inequality symbol is “less than,” so -9 is not included. He should have used an open circle on -9.

Step-by-step explanation:

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The frequency table shows a set of data collected by a doctor for adult patients who were diagnosed with a strain of influenza.

Alexxx [7]

Answer:

The answer would be the second plot

Step-by-step explanation:

The frequency table show how many sick patients there are in a single age range. In the first two plots, the total dots between numbers 24, 25, 26, 27, 28, and 29 add up to 3, and the other 2 plots do not, so those are not correct.

When we add up the total of dots from ages 30-34 on the first plot, there are only 4 sick patients. On the second graph, there are 6 total patients in that age range. Therefore, the dot plot that corresponds with the frequency table is the second one.

7 0

3 years ago

Read 2 more answers

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

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

So, 80% confidence interval for the population mean, $\mu$ is ;

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

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

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

= [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%.

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

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

= [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%.

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

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

= [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

3 years ago

Help please will give brainliest

dimulka [17.4K]

Answer:

the black points on the top right marked with the letters are the answers

Hope it Helps

4 0

2 years ago

Find the first term of a sequence x+1,2x-18,2x-1,if this sequence is an arithmetic progression

PIT_PIT [208]

Answer:

a₁ = 37

Step-by-step explanation:

In an arithmetic progression the common difference d is

d = a₂ - a₁ = a₃ - a₂ , that is

2x - 18 - (x + 1) = 2x - 1 - (2x - 18) ← distribute and simplify both sides

2x - 18 - x - 1 = 2x - 1 - 2x + 18

x - 19 = 17 ( add 19 to both sides )

x = 36

Thus

a₁ = x + 1 = 36 + 1 = 37

6 0

3 years ago

What is 3 1/8 in simplest form?

Deffense [45]

3 1/8 = 25/8

25 and 8 have no common factors, so 25/8 is in simplest form.

6 0

3 years ago