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3 years ago

Hi! I have a geometry question below, could someone please take a look?

Mathematics

1 answer:

AnnyKZ [126]3 years ago

7 0

Answer:

109 degrees.

Step-by-step explanation:

m< AEB = 1/2 (98 + 120)

= 1/2 * 218

= 109 degrees.

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Please I need help in this question, Best answer gets BRAINLIEST

Nata [24]

Answer:

1 - a/variable

2 - d/statistic

3 - b/population

4 - c/data

5 - e/parameter

6 - f/sample

I did comment asking if the two you'd already selected were definitely correct, but no answer was given, so I apologise if this is wrong. This is what I would put.

7 0

4 years ago

Find the product. y4 · y6

Vanyuwa [196]

Answer:

$y^{10}$

Step-by-step explanation:

I am assuming that you mean $y^4*y^6$ .

$y^4*y^6\\\text {Apply the product rule: } a^n*a^m =a^{n+m}\\4 + 6 = 10\\\text {Therefore,}\\\boxed {y^4*y^6 = y^{10}}$

Brainilest Appreciated.

4 0

3 years ago

Read 2 more answers

27? i cant figure out her mistake can anyone help ?

expeople1 [14]

The mess up occurs in one of the first things she says. She first claims that 7^2 +25^2 = 24^2
Wait a minute... slow your roll there.
25 is the hypotenuse, or the longest side of the triangle AND opposite the right angle (but we aren't supposed to know that yet).
The Pythagorean Theorem holds that a^2 + b^2 = c^2, where a & b are the legs of the triangle & c is the hypotenuse. So, she got the formatting correctly, but she switched 24 & 25.
Hope this helps!

4 0

3 years ago

Read 2 more answers

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$

so, $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 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