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2 years ago

4. Solve the following system of equations: y - 3x + 15 = 10 6 - 4x = y

Mathematics

1 answer:

belka [17]2 years ago

5 0

Answer:

erfdewertgfdrtgvcfddddfggggvvvvcerfgvvcddfggvvvvvcddddffffffffffffffffccccccccccccdfvvvvvvv

Step-by-step explanation:

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The quotient of a and b

S_A_V [24]

Answer: take the first digit of a and divide it by b

Step-by-step explanation: and that should give you your answer and if not please feel free to mesage me.

8 0

3 years ago

a 6-pack of undershirts costs 13.98.this is 3.96 less than the cost of buying 6 individual shirts.if each undershirt costs the s

Sedaia [141]

2.99
I got this answer by adding 13.98+3.96=17.94. I divided 17.94÷6 and got 2.99. So, if a undershirt was purchase individualy, it will cost 2.99

7 0

4 years ago

If p(a)=a^3-6a^2+11a-9 and p(a)=-3,find the value of a?

Colt1911 [192]

Correct because if you do the math correctly you solve it

8 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

4 years ago

The tables below show two functions. For each set of data:

slamgirl [31]

The first table, representing f(x), is linear. The data have a constant rate of change or slope:
(between the first two points): m = (y₂ - y₁)/(x₂ - x₁) = (22-18)/(-1--2) = 4/(-1+2) = 4/1 = 4. The rate of change between any two points is the same:
(between the last two points): m = (34-30)/(2-1) = 4/1 = 4.

The second table, representing g(x), is exponential. The data points are multiplied by the same constant between successive points. 2*2 = 4; 4*2= 8; 8*2 = 16, etc.

4 0

3 years ago