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3 years ago

Explain how would you round 25,588 to the nearest ten thousand

2 answers:

8 0

I would look at the biggest number and look to its right..... Then if it is 0-4 stays on the floor 5-9 climbs the vine. It is a 5 so it would round to 26,000

Olin [163]3 years ago

5 0

I would say "The next larger ten thousand is 30,000 ."

Then I would say "The next smaller ten thousand is 20,000 ."

Then I would look at both of them, and see which one is closer to 25,588 .
Whichever one is closer, that would be the nearest ten thousand.

You might be interested in

Which system has no solutions?

seraphim [82]

Answer:

2nd option

y > 6 and y < 2

Step-by-step explanation:

as the value of y can not be greater than 6 and less than 2 at the same time, the 2 inequalities have no common solutions

theefore no solution

8 0

3 years ago

Rachel is a lunchroom supervisor at West School. The children eat lunch at 15 long tables. When all tables are used, 240 childre

Diano4ka-milaya [45]

240÷15=16 seats at each table

Since there were 15 tables and 240 in total you have to divide to find the number of seats at a table

6 0

4 years ago

Read 2 more answers

A clock is positioned on an auditorium wall with its center 9 ft above the floor. The second hand on the clock is 10 inches long

oee [108]

Answer:

B : y=5/6cos(pi/30x)+9

Step-by-step explanation:

Edge 2020

7 0

4 years ago

Read 2 more answers

1/4x+17=x-4 help me and don't answer if you not gonna be serious

iogann1982 [59]

X = 28

1/4+17=x-4
Multiply both sides of the equation by 4
X+68=4z-16
Move variable to the left hand side and change its sign
X-4x+68=-16

X-4x=-16-68
Combine like terms
-3x=-16-68
-3x=-84
Divide both sides by -3
X=28

7 0

3 years ago

Read 2 more answers

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] .

7 0

3 years ago