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Mashutka [201]
3 years ago
11

What is the cost of each of one small box of oranges and one large box of oranges

Mathematics
1 answer:
dezoksy [38]3 years ago
8 0

Answer:The answer would be d, small boxes would cost $4 and

Large boxes would be $10

Step-by-step explanation: multiply the amount of boxes by the cost

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Determine whether the equation 2x + y = 8 is linear. If so, graph the function.
erica [24]

Answer:

Yes, it is a linear equation. This equation is in y=mx+b form, which is a linear equation. To rearrange it into y=mx+b form, you'll just do y= -2x+8

To make a line, you only need two points, so here are two points that fit into the line of y= -2x+8:

(0,8) and (1,6)

8 0
2 years ago
A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week
Levart [38]

Answer:

99% confidence interval for the population proportion of employed individuals is [0.104 , 0.224].

Step-by-step explanation:

We are given that a simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week.

Of the 250 employed individuals​ surveyed, 41 responded that they did work at home at least once per week.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

                              P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of individuals who work at home at least once per week = \frac{41}{250} = 0.164

           n = sample of individuals surveyed = 250

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                             level of significance are -2.5758 & 2.5758}  

P(-2.5758 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<em>99% confidence interval for p</em> = [\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.164-2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } } , 0.164+2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } } ]

 = [0.104 , 0.224]

Therefore, 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is [0.104 , 0.224].

7 0
2 years ago
Need help. I can’t fail this!! 7
vodka [1.7K]

Answer:

Third one? I'm not sure though.

Step-by-step explanation:

3 0
2 years ago
Consider a population with population proportion p, and a sample from the population with sample proportion pˆ. Which of the fol
Deffense [45]

According to CollegeBoard.org:

<u>Answer</u>: "To estimate the probability of observing a value as extreme as  pˆ  given  p."

<u>Explanation</u>: "The test statistic for a one-sample z-test is the distance, in units of standard deviations, between the statistic and the given parameter. From that distance, probabilities (a p-value) can be calculated and a claim can be assessed."

3 0
2 years ago
Help and explanation with this probability question would be really appreciated :)
GuDViN [60]

Answer:

b

Step-by-step explanation:

The probability (PB) is .4 or 4/10 or 2/5.

So, we'd do a proportion.

2/5=x/6

Cross multiply.

5x=12

Divide by sides by 5.

x=2.4

We round up to account for the .4

so b

I can't answer c right now, but I'll come back once my class is over :)

3 0
3 years ago
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