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MrRa [10]
2 years ago
9

Multiply: (2x + 5)(3x + 2) Simplify and write your answer in standard form

Mathematics
2 answers:
Eduardwww [97]2 years ago
4 0

Answer:

The answer is 6x^2+19x+10

Step-by-step explanation:

RideAnS [48]2 years ago
3 0

Answer:

Hey,

\rm \: =6x^2+19x+10

Step-by-step explanation:

=(2x+5)(3x+2)

=(2x)(3x)+(2x)(2)+(5)(3x)+(5)(2)

=6x^2+4x+15x+10

=6x^2+19x+10(ANS)

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5 minus 16 tenths ……
Bess [88]

Answer:

155

Step-by-step explanation:

6 0
1 year ago
The average weight of David, Esther, and Henry is 42.6 kg. Esther is twice as heavy as Henry. David is 5.2 kilograms lighter tha
kari74 [83]

Answer:

  53.2 kg

Step-by-step explanation:

Let E represent Esther's weight. Then Henry's weight is E/2, and David's weight is E-5.2. Their average weight is ...

  ((E-5.2) + E + (E/2))/3 = 42.6

  2.5E -5.2 = 127.8 . . . . . . . . . . multiply by 3, simplify

  2.5E = 133 . . . . . . . . . . . . . . . . add 5.2

  E = 133/2.5 = 53.2 . . . . kg

Esther's weight is 53.2 kg.

7 0
3 years ago
I need to tak to a female beecause i have a few questions
NikAS [45]

Answer:

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7 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

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

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

<u><em>95% confidence interval for</em></u> \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].

<em>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.</em>

8 0
3 years ago
Are 5(2x-y) and 15x-5y equivalent expressions?
krek1111 [17]

No, they are not equivalent.

5(2x - y) = 10x - 5y and it is not equal to 15x - 5y.

(using distribution property)

6 0
3 years ago
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