Find the area of the figure. PLS I NEED HELP FAST I BEG YOU!

Please help me :( I really need it.

Genrish500 [490]

Answer:

6

Step-by-step explanation:

since it's a right triangle you can use the Pythagorean Theorem

2.7² + b² = 6.5²

7.29 + b² = 42.25

b² = 34.96

b = $\sqrt{34.96}$ which equals 6 when rounded to the nearest tenth

6 0

2 years ago

Which linear equation represents the graph?

8090 [49]

Answer:

The answer is C. Y= -3x+6

The answer can be found by substitute 0 into x and find the value of y. If the value is same with graph then it is correct.

Thank you and please rate me as brainliest as it will help me to level up

4 0

2 years ago

Expand 5(2x - 1) plsssss help some one

diamong [38]

10x-5

You multiply each term in the parentheses by 5

3 0

2 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

Dishwashers are on sale for 25% off the original price ( d ), which can be expressed with the function p ( d ) = 0.75d. Local ta

Marta_Voda [28]

Price of the dishwashers on sale are represented with the function: p ( d ) = 0.75 d ( 100 % - 25 % = 75 % or 0.75 d ): After that we have additional 14 % on the discounted price ( 100 % + 14 % = 114 % OR 1.14 ): C ( P ) = 1.14 P. Finally, c ( p ( d ) ) = 1.14 * 0.75 d = 0.855 d. Answer: C ) c [ p ( d ) ] = 0.855 d Hope I helped! :) Cheers!

8 0

3 years ago