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Bumek [7]
3 years ago
12

Veronica bought 20 bags of candy for a school dance the first 5 bags cost $1.79 each the rest of the bags cost $1.19 each how mu

ch did veronica spend on candy
Mathematics
1 answer:
Stella [2.4K]3 years ago
5 0
Answer: Veronica spent $26.8

Explanation:

5 x $1.79= $8.95

15 x $1.19= $17.85

$8.95 + $17.85= $26.8
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Synonym for precious
NikAS [45]

Answer:

important, valuable, nice,

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3 years ago
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Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t
musickatia [10]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

8 0
3 years ago
Find the sum of 7g and 4g + 2
Shtirlitz [24]

Answer:

13g

Step-by-step explanation:

7g+4g+2=

11g+2=

13g

5 0
2 years ago
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Two shipping containers are being loaded into the cargo hold
Nesterboy [21]

Answer:

1.85 Tons

Step-by-step explanation:

4.15 minus 2.3 equals 1.85

5 0
3 years ago
Silk scarves are produced in a factory. Quality control investigations find that 0.5% of the scarves produced have flaws, which
Goshia [24]

The binomial probability is determined by the formula

\begin{gathered} P_x=\binom{n}{x}p^xq^{n-x} \\ \text{where} \\ P\text{ is the binomial probability} \\ x\text{ is the number of times for a specific outcome within }n\text{ trials} \\ p\text{ is the probability of success on a single trial} \\ q\text{ is the probability of failure on a single trial} \\ n\text{ is the number of trials} \end{gathered}

Here, we will assume that finding the flaw is the "success" of the trial hence the following given

\begin{gathered} p=0.05 \\ q=0.95 \\ n=30 \end{gathered}

Finding the probability of one flawed scarf

\begin{gathered} P(X=1)=\binom{30}{1}(0.05)^1(0.95)^{30-1} \\ P(X=1)=0.3389 \end{gathered}

Finding the probability of no flawed scarves

\begin{gathered} P(X=0)=\binom{30}{0}(0.05)^0(0.95)^{30-0} \\ P(X=0)=0.2146 \end{gathered}

Finding the probability of more then three flawed scarves

\begin{gathered} P(X>3)=1-P(X\le3) \\ P(X>3)=1-0.93922843869 \\ P(X>3)=0.0608 \end{gathered}

3 0
1 year ago
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