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3 years ago

Write a 4 -digit number that is divisible by both 5 and 9

Mathematics

1 answer:

vodomira [7]3 years ago

6 0

4500 is both divisible by 5 and 9
4500/9=500
4500/5=900

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The last answer is D. 15 meters

Lapatulllka [165]

5^2 +12^2= diagonal^2
25+144=d^2
169= d^2
D= 13

4 0

3 years ago

What is the decimal equivalent of 17/20? A. 0.085 B. 0.117 C. 0.85 D. 8.5

RideAnS [48]

I believe it’s the answer is C

5 0

2 years ago

Read 2 more answers

-3y = 2x - 9; x= -3, 0, 3

astraxan [27]

1.)-3y=2* -3-9
  -3y=-6-9
  -3y=-15
  y=5
-3*5=-15
2.)-3y=2*0-9
-3y=0-9
-3y=-9
y=3
-3*3=-9
3.)-3y=2*3-9
-3y=6-9
-3y=-3
y=1
-3*1=-3

3 0

3 years ago

Jay has 14 blocks in all he has6 yellow blocks the rest of the blocks are green how many green blocks does jay have

-Dominant- [34]

Answer:

To find this out you would take the amount of yellow bricks minus the amount of bricks in total

14 blocks - 6 yellow blocks = 8 blocks

from this information we can conclude that jay has 8 green blocks

Step-by-step explanation:

4 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

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 mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago