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3 years ago

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:

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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

3 years ago

Please help with this problem with steps

Hoochie [10]

x^4 - 1
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= 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