All categories

2 years ago

Which expression is the same as 5 + 3?

Mathematics

1 answer:

dimaraw [331]2 years ago

5 0

Answer:

Step-by-step explanation:

5 + 3 = 8

Hope that helps!

-Sabrina

You might be interested in

PLZ HURRY IT'S URGENT!!!

mr Goodwill [35]

Answer:

25 two point shots

25 three point shots

2x+3y<100 is that the one missing the sign?

Step-by-step explanation:

25*2=50

25*3= 75

50+75= 125

2x+3y has to be over 100 points

so 2x+3y<100

8 0

3 years ago

The sum of a number X and four all divided by three

sergey [27]

If you are looking for the expression, that would be (x+4)/3

4 0

2 years ago

Read 2 more answers

Please help if you can

arsen [322]

Alright what is VAT?

5 0

3 years ago

Laura is baking a cake. The recipe says that she has to mix 64 grams of chocolate powder to the flour. Laura knows that 1 cup of

pickupchik [31]

The answer is She went over by one over six of a cup

1 cup of that particular chocolate powder has a mass of 128 grams: 1c = 128g

64 grams of chocolate powder is x cups.
1c = 128g
x = 64g

1c : 128 g = x : 64g
x = 1c : 128 g * 64 g
x = 0.5 c
64 grams of chocolate powder is 0.5 cups = 1/2 cups

She added two over three of a cup of chocolate powder: 2/3 cups

She need to add 1/2 cups: 1/2 = 3/6 cups
She added 2/3 cups: 2/3 = 4/6 cups

So she added 4/6 - 3/6 = 1/6 cups more.

5 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