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yan [13]
3 years ago
7

Joes account was 1 more than one half of Chandras account. The sum of their accounts was $751. Find the amount of Joes account

Mathematics
1 answer:
Minchanka [31]3 years ago
4 0

Answer:

251$

Step-by-step explanation:

Let's say x represents Joes's account.

And y represents Chandra.

Since Joses account was 1 more than one half of chandra, the amount for Joes would be x = 1/2*y + 1

Now we can just say y = Chandra's amount.

So x + y = 751

And x = 1/2*y + 1

We have a systems of equation, and just plug in 1/2*y + 1 for x.

1/2*y + 1 + y = 751

1/2*y + y = 750 (subtract both sides by 1)

distribute y so...

y(1/2 + 1) = 750

y = 750/1.5

y = 500

Remember, y is Chandra. So plug in 500 for y in Joses equation, so 1/2(500) +

is 250 +1 so 251.

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A simple random sample of size n equals 200 individuals who are currently employed is asked if they work at home at least once p
____ [38]

Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

Step-by-step explanation:

<u>step 1:-</u>

Given sample  size n=200

of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week

 Population proportion of employed individuals who work at home at least once per week  P = \frac{x}{n} =\frac{41}{200} =0.205

Q=1-P= 1-0.205 = 0.705

<u>step 2:-</u>

Now  \sqrt{\frac{P Q}{n} } =\sqrt{\frac{(0.205)(0.705)}{200} }

=0.0015

<u>step 3:-</u>

<u>Confidence intervals</u>

<u>using formula</u>

(P  -  Z_∝} \sqrt{\frac{P Q}{n},} (P  +  Z_∝} \sqrt{\frac{P Q}{n},

(0.205-2.58(0.0015),0.205+2.58(0.0015)\\0.20113,0.20887

=0.20113,0.20887[/tex]

<u>conclusion:</u>-

99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

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