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2 years ago

1.) if a person weighs 160 pounds, what is his mass in kilograms

Mathematics

1 answer:

Stella [2.4K]2 years ago

4 0

Answer:

72.56

Step-by-step explanation:

Divide the number of pounds by 2.205

160/2.205 = 72.56

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If the reserve rate is 7% and a bank receives a deposit of $9000, how much of the $9000 is the bank free to lend?

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3 years ago

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2 years ago

A sprinkler system is being installed in a newly renovated building on campus. The average activation time is supposed to be at

melomori [17]

Answer:

A) t = 1.73

B) p-value = 0.0558

C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05

D) The decision means that the design specifications are not met.

E) Type II error

Step-by-step explanation:

The hypotheses are:

H₀: μ = 20

H₁: μ > 20

A) Formula for the test statistic is;

t = (x' - μ)/(s/√n)

Now, we are given;

x' = 21.5

μ = 20

s = 3

n = 12

Thus;

t = (21.5 - 20)/(3/√12)

t = 1.73

B) we have our t-value as 1.73

Now, Degree of freedom(DF) = n - 1

So,DF = 12 - 1 = 11

Using significance level of α = 0.05, t-value = 1.73 and DF = 11, one tailed hypothesis, from online P-value calculator attached, we have;

p-value = 0.0558

C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05

D) We will not reject the null hypothesis. The decision means that the design specifications are not met.

E) If the true average activation time of the sprinkler system is, in fact, equal to 20 seconds, then the null hypothesis is false.

Since we did not reject the null hypothesis even though it is false, the error that was committed was therefore a type II error.

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3 years ago

Larry bought several boxes of baseball cards at a yard sale and ended up with 204

AysviL [449]

Answer:

Step-by-step explanation:

204/4=51

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3 years ago

Read 2 more answers

A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen

tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average desired retirement age = 55 years

$\sigma$ = sample standard deviation = 3.4 years

n = sample of seniors = 101

$\mu$ = true mean retirement age of all college students

Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 96% confidence interval for the population mean, $\mu$ is ;

P(-2.114 < $t_1_0_0$ < 2.114) = 0.96 {As the critical value of t at 100 degree

of freedom are -2.114 & 2.114 with P = 2%}

P(-2.114 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.114) = 0.96

P( $-2.114 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $55-2.114 \times {\frac{3.4}{\sqrt{101} } }$ , $55+2.114 \times {\frac{3.4}{\sqrt{101} } }$ ]

= [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

7 0

3 years ago

1.) if a person weighs 160 pounds, what is his mass in kilograms ​

1.) if a person weighs 160 pounds, what is his mass in kilograms