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joja [24]
3 years ago
15

The total charge on 7 particles is −42 units. All the particles have the same charge.

Mathematics
2 answers:
WITCHER [35]3 years ago
8 0

Your answer would be -6! -42= -6 divided by 7 or 7 X -6= -42

qwelly [4]3 years ago
5 0
If they have the same charge
x+x+x+x+x+x+x=-42
7x=-42
x=-6
so to check our answer
-6-6-6-6-6-6-6=-42
-12-12-12-6=-42
-24-18=-42
-42=-42
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8 times the difference of a number and 3 is equal to the number increased by 25. what is the number?
madam [21]

Answer:

the answer is four

The solution is in the picture above please mark me brainliest :)

3 0
3 years ago
A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week
Levart [38]

Answer:

99% confidence interval for the population proportion of employed individuals is [0.104 , 0.224].

Step-by-step explanation:

We are given that a simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week.

Of the 250 employed individuals​ surveyed, 41 responded that they did work at home at least once per week.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

                              P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of individuals who work at home at least once per week = \frac{41}{250} = 0.164

           n = sample of individuals surveyed = 250

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                             level of significance are -2.5758 & 2.5758}  

P(-2.5758 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<em>99% confidence interval for p</em> = [\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.164-2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } } , 0.164+2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } } ]

 = [0.104 , 0.224]

Therefore, 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is [0.104 , 0.224].

7 0
2 years ago
Please help with this problem with steps
Hoochie [10]
   x^4 - 1
-------------
   x^2 + 1

    (x^2 +1) (x^2 - 1)
= ------------------------
           x^2 + 1

 = x^2 - 1 ..................this is your answer
 

3 0
3 years ago
The frog population at a lake doubles every week. The population can be modeled by f(x) = 15(2)^x and f(6) = 960. What does the
Svetach [21]

we have f(x)=15.2^x we have f(6)=960

It means we at x=6 the population of frog is 960 where the x is meant for sixth week .

so the correct option is D )the number of weeks that have passed

6 0
3 years ago
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Taya2010 [7]

Answer:

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10C

6 0
3 years ago
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