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arsen [322]
4 years ago
9

What is sum of even number from 1 to 1000??

Mathematics
2 answers:
babymother [125]4 years ago
7 0
The sum of all the even numbers between 1 and 1000 is 2550
Mekhanik [1.2K]4 years ago
5 0
I would say there are 500 even numbers and 500 odd numbers so I would say 500 is the sum
You might be interested in
Looking for some help!
klasskru [66]

Answer:

  • 5√2
  • 16

Step-by-step explanation:

<u>Use the property of 45 degree right triangles</u>

  • c = a√2, where c- hypotenuse, a = legs

<u>Top diagram</u>

  • x = 5√2

<u>Bottom diagram</u>

  • x = 8√2(√2) = 16

5 0
3 years ago
Find the length of BC.<br><br><br><br>Explain how you got it, please!<br>Thanks!
Vinvika [58]

Answer:

BC = 30.73

Here,

\sf \frac{AB}{BQ}  = \frac{CD}{DQ}

so first solve for QD

\sf \hookrightarrow \frac{32}{15}  = \frac{19.2}{DQ}

\sf \hookrightarrow 32(DQ)}  =19.2(15)

\sf \hookrightarrow 32(DQ)}  =288

\sf \hookrightarrow DQ =9

  • Hence, QD = 9

Now! <u>using Pythagoras theorem,</u>

  • CD² + BD² = BC²
  • 19.2² + (9+15)² = BC²
  • BC = √368.64+576
  • BC = 30.73499634
  • BC = 30.73 ( rounded to nearest hundredth )
4 0
2 years ago
Read 2 more answers
A jewelery store paid $75 for a necklace at wholesale cost. The store will sell the necklace for $250. What is the percente mark
pickupchik [31]
40 percent thanks for using Brainly
6 0
3 years ago
Read 2 more answers
It takes 50 oz of grass seed to seed 3000 ft2 of lawn. At this​ rate, how much would be needed for 7200 ft2 of​ lawn?
saw5 [17]

Answer:

120 ounces

Step-by-step explanation:

First find how many ft2 of lawn 1 once can seed.

To do this we simply divide 3000 by 50 ( because 50 ounces of grass seed can seed 3000 ft2 of lawn )

3000 / 50 = 60

So 1 ounce can seed 60ft2 of lawn

To find how many ounces would be needed to seed 7200 ft2 of lawn we would divide 7200 by the amount of lawn 1 once can seed ( which is 60ft2 )

7200/60 = 120

So 120 ounces of grass seed would be needed to seed 7200 ft2 of lawn

6 0
2 years ago
Kids with cell phones: A marketing manager for a cell phone company claims that less than 58% of children aged 8-12 have cell ph
Vikentia [17]

Answer:

z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32  

p_v =P(z  

Since the p value is lower than the significance level \alpha=0.1 we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.

For the Ti84 preocedure we need to do this:

STAT> TESTS> 1-Z prop Test

And then we need to input the following values:

po= 0.58

x = 443 , n= 820

prop <po

And then calculate and we will get the same results

Step-by-step explanation:

Data given and notation

n=820 represent the random sample taken

X=443 represent the people that had cell phones

\hat p=\frac{443}{820}=0.540 estimated proportion of people that had cell phones

p_o=0.58 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.58.:  

Null hypothesis:p\geq 0.58  

Alternative hypothesis:p < 0.58  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

And replacing we got:

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32  

Decision

The significance level provided \alpha=0.1. Now we can calculate the p value

Since is a left tailed test the p value would be:  

p_v =P(z  

Since the p value is lower than the significance level \alpha=0.1 we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.

For the Ti84 preocedure we need to do this:

STAT> TESTS> 1-Z prop Test

And then we need to input the following values:

po= 0.58

x = 443 , n= 820

prop <po

And then calculate and we will get the same results

5 0
3 years ago
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