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3 years ago

7

Dennis mowed his next door neighbor's lawn for a handful of dimes and nickels 80 coins in all. upon completing the job he counte

d out the coins and it came to $6.60 how many of each coin did he earn?

1 answer:

lbvjy [14]3 years ago

6 0

54 Dimes and 28 nickles

You might be interested in

Shannon was born on 04/02/2004. How many eight-digit codes could she make using the digits in her birthday?

LuckyWell [14K]

Answer:

24 eight-digit code.

Step-by-step explanation:

Frequency : 0 : 4

2 : 2

4 : 2

In total there are Eight (8) numbers and 3 frequent number ( which are 0, 2, 4 )

∴ 8 × 3 = 24.

6 0

3 years ago

Day people The chart shows how many people have signed up to go on a field trip each day. How many people would you expect to si

alexandr402 [8]

D) 50 because the chart is increasing by 4 each time.

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N111ancy out!

8 0

3 years ago

Read 2 more answers

The probability that a grader will make a marking error on any particular question of a multiple-choice exam is 0.15. If there a

sashaice [31]

Answer:

$P(X=0)=(10C10)(0.15)^{0} (1-0.15)^{10-0}=0.1969$

$P(X \geq 1)= 1-P(X$

$P(X=0)=(nCn)(p)^{0} (1-p)^{n-0}=(1-p)^n$

$P(X \geq 1)= 1-P(X$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.15)$

what is the probability that no errors are made?

For this case means that all the questions were correct so we want this probability:

$P(X=0)=(10C10)(0.15)^{0} (1-0.15)^{10-0}=0.1969$

what is the probability that at least one error made?

For this case we want this probability:

$P(X \geq 1)$

And we can use the complement rule:

$P(X \geq 1)= 1-P(X$

If there are n questions and the probability of a marking error is p rather than 0.15, give expressions for the probabilities of no errors and at least one error

$P(X=0)=(nCn)(p)^{0} (1-p)^{n-0}=(1-p)^n$

$P(X \geq 1)= 1-P(X$

8 0

3 years ago

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

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

<u>Part I</u> : 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

<u />

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

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

= [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].

<u>Part II</u> : 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

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

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

<u />

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

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

= [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

3 years ago

Find the missing end point if H is the midpoint of PS. P(2,5) and H (3,4) find S (endpoint)

RUDIKE [14]

Answer:

(4,3)

Step-by-step explanation:

To find the midpoint add the end points and divide by 2

P(2,5) S ( sx, sy) and the midpoint is and H (3,4)

X coordinate

( 2+ sx) /2 = 3

Multiply each side by 2

2+sx = 3*2

2+sx = 6

Subtract 2

sx = 4

Y coordinate

( 5+ sy) /2 = 4

Multiply each side by 2

5+sy = 4*2

5+sy = 8

Subtract 5

sy = 3

The end point is

(4,3)

8 0

3 years ago

Read 2 more answers