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3 years ago

A football player lost 10 yards when he was tackled. a 10 b -10

Mathematics

2 answers:

PSYCHO15rus [73]3 years ago

6 0

B. -10 because the football player “lost” implies a reduction in yards

Ganezh [65]3 years ago

3 0

The key word “lost” indicates that the football player went back 10 yards. The correct answer is B. -10! :)

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What are the roots of the function f(x) = x2 + 2x – 1? a. x = ± √ _ 2 b. x = − 1 ± √ _ 2 c. x = 2 ± √ _ 2 d. x = 1 ± √ _ 2

Masteriza [31]

Step-by-step explanation:

Root of the function, where y = 0

f(x) = x² + 2x - 1

0 = x² + 2x - 1

x² + 2x - 1 = 0

x² + 2x = 1

x² + 2x + 1 = 1 + 1

(x + 1)² = 2

x + 1 = ±√2

x = -1 ± √2

Roots → x = -1 ± √2

7 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

Henry is mixing concrete. To make concrete he needs to mix cement, sand and gravel in the ratio 1 : 4 : 7 by weight. Henry needs

jolli1 [7]

Answer:

Henry is short of gravel

Step-by-step explanation:

Here, we are to identify which of the ingredients Henry is short of.

To identify this ingredient, what is needed to be done is to share the 180kg of concrete appropriately to get which of the material does not meet its ratio and thus allowing us know which is insufficient.

The total ratio is 1 + 4 + 7 = 12

For cement, we have ;

1/12 * 180 = 15 kg

For sand, we have

4/12 * 180 = 60kg

For gravel, we have

7/12 * 180 = 105 kg

From the values in the question, we can see that all the values we have are greater what is shared in the ratio except in the case of gravel. So we can confidently say he is short of gravel

8 0

3 years ago

PLEASE HELP!! this is due soon and i rlly need help

egoroff_w [7]

Answer:

Solution given:

r=3km

area of sector bounded by a 90° arc=90°/360°*πr²

=¼*π*3²

=9/4 π or 7.068km²

is a required answer

6 0

2 years ago

Read 2 more answers

Case Study 2

ICE Princess25 [194]

Answer:

good for them i think

Step-by-step explanation:

this doesnt make any sense

4 0

2 years ago