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2 years ago

I need help please someone explain

Mathematics

2 answers:

avanturin [10]2 years ago

8 0

Answer:

Together they both hiked $1\frac{1}{4}$ miles

Step-by-step explanation:

Dylan hiked $\frac{3}{4}$ of a mile and Sonia hiked $\frac{1}{2}$ of a mile or $\frac{2}{4}$ of a mile. Add both together and you get $\frac{5}{4}$ or $1\frac{1}{4}$

masha68 [24]2 years ago

3 0

Answer: Dylan Hiked 1.25 miles and Sonia Hiked 1.35 miles

Step-by-step explanation:

Dylan you add 3/4+1/2

Sonia you add 1/2+3/5

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Find all the real cube roots of 1000/27 10/3 10/9 -10/9 -10/3

gulaghasi [49]

Can you do prime factoriztion

1000
100 * 10
10 * 10 * 10

27
9 * 3
3 * 3 * 3

cube root (1000/27) = 10/3

7 0

2 years ago

Read 2 more answers

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago

Mr. Jones eats out at his favorite buffet restaurant every Friday night. If the cost is $9 and he budgets $42 for the month, how

alexandr402 [8]

Answer:

He will be able to eat nine times.

Step-by-step explanation:

Because nine times nine is 91 which is on one less then what he has, so that means he will be able to eat there nine times. Hope I helped

7 0

2 years ago

1+ 3z= 40 what’s the answer to this problem I’m confused???

stepan [7]