Determine the measure of angle j: angle k: & angle: l

54% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

Liula [17]

Answer:

$P(5)=0.238$

$P(x\geq 6)=0.478$

$P(x$

Step-by-step explanation:

In this case we can calculate the probability using the binomial probability formula

$P(X=x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}$

Where p is the probability of obtaining a "favorable outcome " x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.

(a) exactly five

This is:

$x=5,\ n=10,\ p=0.54.$

So:

$P(X=5)=\frac{10!}{5!(10-5)!}*0.54^x*(1-0.54)^{10-5}$

$P(5)=0.238$

(b) at least six

This is: $x\geq 6,\ n=10,\ p=0.54.$

$P(x\geq 6)=P(6) + P(7)+P(8)+P(9) + P(10)$

$P(x\geq 6)=0.478$

(c) less than four

This is: $x< 4,\ n=10,\ p=0.54.$

$P(x$

8 0

3 years ago

Read 2 more answers

Simplify: (2x2 − 9x + 3) + (−7x2 + 4x − 2)

Leokris [45]

Combine like terms
2x^2 + -7x^2 = -5x^2
-9x + 4x = -5x
3 + -2 = 1
put back together
-5x^2 - 5x +1

4 0

2 years ago

List all possible rational zeros of h(x) = x^3 - 5x^2 - 4x + 20. Then determine which, if any, are zeros.

devlian [24]

Answer: x = {2, 5, -2}

Explanation:

see attachment

4 0

2 years ago

Help find the missing segment???

SVEN [57.7K]

(9/6) = (15/?)
9? = 90
? = 10

6 0

2 years ago

Read 2 more answers

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% 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 nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% 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 nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

2 years ago