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3 years ago

I am an odd 4-digit number between 5,000 and 5,999. The digit in the thousands place is half

1 answer:

7 0

You are 5,987. 5 in the thousands place because you are a number in the 5,000s, and also the smallest 2 digit number divided 2 makes 5. 9 in the hundreds place because the biggest 1 digit number is 9. 8 in the tens place because the only single digit number that is a multiple of 4, but not 4, is 8. 7 in the ones place because 9 is the hundreds place, minus 2 is 7.

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What is the answer for 2 5/7 - 7/8

enot [183]

Change the mixed leading number into an improper fraction...a b/c=(ac+b)/c

19/7-7/8 the least common multiple of the denominators is 56 so:

(19/7)(8/8)-(7/8)(7/7)

(152-49)/56

103/56 now we can change back into a mixed number...

(56+47)/56

1 47/56

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3 years ago

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2x+x(x+6) Solve to the lowest equation/number

Ghella [55]

Answer:

x^2 +8x

Step-by-step explanation:

2x+x(x+6)

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2x + x^2 +6x

Combine like terms

x^2 +8x

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Bas_tet [7]

Answer:

Step-by-step explanation:

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3 years ago

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If a pound of apples cost $2.90 how much would 2.4 lbs cost

polet [3.4K]

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3 years ago

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An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y

aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

Firstly, the pivotal quantity for finding the 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 time = 23.92 minutes

s = sample standard deviation = 6.72 minutes

n = sample of times = 40

$\mu$ = population mean assembly time

Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, a 98% confidence interval for the population mean, $\mu$ is;

P(-2.426 < $t_3_9$ < 2.426) = 0.98 {As the critical value of z at 1% level

of significance are -2.426 & 2.426}

P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

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3 years ago