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guajiro [1.7K]
3 years ago
13

Please help if you can:)

Mathematics
1 answer:
geniusboy [140]3 years ago
8 0

Answer:

(-2,17) i think i am not sure

Step-by-step explanation:

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A juice drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of drink on an average. Any overfill
ozzi

Answer:

d. The average is equal to 12 ounces.

Step-by-step explanation:

In this problem, the drink filling machine must be perfectly calibrated at 12 ounces since it needs to be shut down in cases of overfilling (mean > 12 ounces) and underfilling (mean < 12 ounces). Therefore, the correct approach would be to test if the mean is 12 ounces and the correct set of hypothesis would be:

H_0:\mu=12\\H_a:\mu\neq 12

The correct alternative is d. The average is equal to 12 ounces.

5 0
3 years ago
A paint mix uses red and white paintin a ratio of 1:12.
levacccp [35]

Answer:

(a).16.8 litres

(b).red-1.2 litres

white-14.3 litres

7 0
2 years ago
Solve for x. Enter the solutions from least to greatest.
Len [333]

Answer:

This is the answer of your question ☺☺

Step-by-step explanation:

(x+7)2 -49=0

2x+14-49=0

2x-35=0

2x=35

x=17.5

7 0
2 years ago
Blood pressure: High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. a
Y_Kistochka [10]

Answer:

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

For this case we know this:

n=37 ,  p=0.2

We can find the standard error like this:

SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658

So then our random variable can be described as:

p \sim N(0.2, 0.0658)

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

P(p>0.4)

We can use the z score given by:

z = \frac{p -\mu_p}{SE_p}

And using this we got this:

P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

Step-by-step explanation:

We need to check if we can use the normal approximation:

np = 37 *0.2 = 7.4 \geq 5

n(1-p) = 37*0.8 = 29.6\geq 5

We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

For this case we know this:

n=37 ,  p=0.2

We can find the standard error like this:

SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658

So then our random variable can be described as:

p \sim N(0.2, 0.0658)

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

P(p>0.4)

We can use the z score given by:

z = \frac{p -\mu_p}{SE_p}

And using this we got this:

P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

7 0
3 years ago
In which question is r = -17 a solution?
asambeis [7]

Answer:

None

Step-by-step explanation:

For option A

3-2r/4=10 and combining like terms and re-arranging we obtain

-2r/4=10-3

-2r/4=7

r=7*4/-2=-14

For option B

3/4-2r=10 then combining like terms and re-arranging

-2r=10-3/4

-2r=9 1/4

r=(9 1/4)/-2=-4.625

For option C

3(r-4)=10 dividing both sides by three first

r-4=10/3

r=10/3+4= 7 1/3

For option D

3-r/4=10

-r/4=10-3

-r/4=7

r=7*-4=-28

Therefore, none of the options yields r=-17

6 0
3 years ago
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