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1 year ago

Brayden was given a box of assorted chocolates for his birthday. Each night, Brayden

Mathematics

1 answer:

Anestetic [448]1 year ago

6 0

Answer:

Step-by-step explanation:

The slope of the function is -3 which reveals the number of chocolates Brayden eats each night.

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How does this make you feel?

Aleks [24]

Answer:

sad to be honest

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Please help me again

Vikentia [17]

Answer:

y = 10

Step-by-step explanation:

Since all sides are equal, that means all angles are 60 degrees. 60 + 60 + 60 = 180

5y + 10 - 10 = 60 - 10

5y / 5 = 50 / 5

y = 10

Answer: y = 10

7 0

3 years ago

An airplane flies due north from ikeja airport for 500km.It then flies on a bearing of 060 from a further distance of 300km befo

Nata [24]

Answer:

482 km

63.94 degrees

Step-by-step explanation:

to solve this question we will use the cosine rule. For starters, draw your diagram. From point A, up north is 500km and 060 from there, another 300. If you join the point from the road junction back to the starting point, yoou have a triangle.

Cosine rule states that

C = $\sqrt{A^{2} + B^{2} -2AB cos(c) }$

where both A and B are the given distances, 500 and 300 respectively, C is the 3rd distance we're looking for and c is the given angle, 060

solving now, we have

C = $\sqrt{500^{2} + 300^{2} -2 * 500 * 300 cos(60) }$

C = $\sqrt{250000 + 90000 - [215000 cos(60) }]$

C = $\sqrt{340000 - [215000 * 0.5 }]$

C = $\sqrt{340000 - [107500 }]$

C = $\sqrt{232500}$

C = 482 km

The bearing can be gotten by using the Sine Rule.

$\frac{sina}{A}$ = $\frac{sinc}{C}$

sina/500 = sin60/482

482 sina = 500 sin60

sina = $\frac{500 sin60}{482}$

sina = 0.8983

a = sin^-1(0.8983)

a = 63.94 degrees

6 0

3 years ago

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

Use a graphing calculator to determine which equation for the line of regression, Pearson product-moment correlation value (r),

Ipatiy [6.2K]

The best answer is:
B) y = 6.1x + 74.5, r = 0.48
Predicted test score for 4 hours of study equals 99.
See the attachment for graph

6 0

3 years ago