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anyanavicka [17]

3 years ago

13

Kathi and Robert Hawn had a pottery stand at the annual Skippack Craft Fair. They sold some of their pottery at the original pri

ce of $9.50 each, but later decreased the price of each by
$3.00. If they sold all 95 pieces and took in $707.50, find how many they sold at the original price and how many they sold at the reduced price

The number sold at $9.50 is

The number sold at $6.50 is

1 answer:

Tanzania [10]3 years ago

3 0

Answer:

Amount of potteries sold at $9.50 is 30

And that of $6.50 is 65 potteries

Step-by-step explanation:

Kindly refer to the attached image for explanations

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ladessa [460]

Combine like terms
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3 years ago

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HELP ASAP 10 POINTSSS

kiruha [24]

Answer:

Complementary angles, adjacent angles

Step-by-step explanation:

Which relationships describe angles 1 and 2?

Options: Complementary angles, vertical angles, supplementary angles, adjacent angles

Angles 1 and 2 are opposite to the 90° angle. So, the angles' relationship to that angles is a vertical angle, meaning that the two add up to 90°.

Complementary angles add up to 90°, so this option is correct.

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I hope this helps! Feel free to ask any questions!

6 0

2 years ago

Help me please I really need!!! :)

WITCHER [35]

Answer:

$\huge\boxed{\sf x = 1.8}$

Step-by-step explanation:

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5 0

3 years ago

Because you are working a late shift, you will get an increase in your wage. You will get and additional 10% per hour from 6:00p

siniylev [52]

CHECK THE COMPLETE QUESTION BELOW:

Supervisor: "Because you are working a late shift, you will get an increase in your wage. You will get an additional 10% per hour from 6:00 PM to 12:00 AM and additional 15% from 12:00 AM to 1:00 AM." Employee: "I am working from 2:00 PM to 12:00 AM. Between 2:00 PM and 6:00 PM, I will make $12.00 per hour. Between 6:00 PM and 12:00 AM, I will make __________ per hour

Answer:

I will make $13.2 per hour

Explanation:

It can be deducted from the question that that the money Employee gets money for Working from 2.00 PM to 6.00 PM = $12per hour

Also the Employee gets additional 10% of money for working from 6.00 PM to 12.00 AM which implies that

10% = (10/100)*$12per hour

=$1.2per hour

Again, Employee gets additional 15% of the money for working from 12.00 AM to 01.00 AM which can be calculated as:

15% = (15/100)*$12per hour

=$1.8 per hour

which means that Employee gets additional $1.8 per hour

of the money for working from 12.00 AM to 01.00 AM

Therefore, if Employee works from 2.00 PM to 12.00 AM,

Employee will gets money for Working from 2.00 PM to 6.00 PM as well as the 10\% extra wage which

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7 0

3 years ago

Suppose a production line operates with a mean filling weight of 16 ounces per container. Since over- or under-filling can be da

miss Akunina [59]

Answer:

From the question we are told that

The population mean is $\mu = 16$

The sample size is n = 30

The sample mean is $\= x = 16.32$

The population standard deviation is $\sigma = 0.8$

The level of significance is $\alpha = 0.10$

Step 1: State hypotheses:

The null hypothesis is $H_o : \mu = 16$

The alternative hypothesis is $H_a : \mu \ne 16$

Step 2: State the test statistic. Since we know the population standard deviation and the sample is large our test statistics is

$t = \frac{ \= x -\mu }{ \frac{\sigma}{ \sqrt{n} } }$

=> $t = \frac{ 16.32 -16 }{ \frac{0.8 }{ \sqrt{30} } }$

=> $t =2.191$

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 30 - 1$

=> $df =29$

Step 3: State the critical region(s):

From the student t-distribution table the critical value corresponding to $\alpha = 0.10$ is

$t = 1.311$

Generally the critical regions is mathematically represented as

$- 1.311 < T < 1.311$

Step 4: Conduct the experiment/study:

Generally the from the value obtained we see that the t value is outside the critical region so the decision is [Reject the null hypothesis ]

Step 5: Reach conclusions and state in English:

There is sufficient evidence to show that the filling weight has to be adjusted

Step 6: Calculate the p-value associated with this test. How does this the p-value support your conclusions in Step 5?

From the student t-distribution table the probability value to the right corresponding to $t =2.191$ at a degree of freedom of $df =29$ is

$P( t > 2.191) = 0.0183$

Generally the p-value is mathematically represented as

$p-value = 2 * P( t > 2.191 )$

=> $p-value = 2 * 0.0183$

=> $p-value = 0.0366$

Generally looking at the value obtained we see that $p- value < \alpha$ hence

The decision rule is

Reject the null hypothesis

Step-by-step explanation:

4 0

3 years ago