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3 years ago

Please help and show work thank you!

Mathematics

2 answers:

MariettaO [177]3 years ago

7 0

Answer:

1)-0.65 2) 28.8 3) 747.6

Step-by-step explanation:

divide the numerator (-13) and the denominator (20)
32 x 90= 2880, divide 2880 by 100 equals to 28.8

3. multiply 12 by 62.3

I hope this helps!

oksano4ka [1.4K]3 years ago

6 0

The is tkdb rndnrndnrndnrjdbr r

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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

Lana sells lampshades for $8 each and paper lanterns for $5 each at a one-day craft fair. She sells k lampshades and (k+5) paper

DENIUS [597]

Answer:

8k + 5k + 25

13k + 25

Step-by-step explanation:

Lana's total sales = 8k + 5(k+5)

Since the expressions are not given in the question;

The equivalent expression could be

= 8k + 5k + 25

Equivalent to

= 13k + 25

3 0

3 years ago

PLEASE HELP ME!! THANKS!! BRAINLIEST IF CORRECT!

Murrr4er [49]

When the calcium atom loses electrons, it's charge increases by the number of electrons lost.

It loses 2 electrons, so it's charge increases by 2.

The answer is +2

4 0

3 years ago

Read 2 more answers

Franchise Business Review stated over 50% of all food franchises earn a profit of less than $50,000 a year. In a sample of 130 c

nydimaria [60]

Answer:

We need a sample size of 564.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$\pi = \frac{81}{130} = 0.6231$

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Based upon a 95% confidence interval with a desired margin of error of .04, determine a sample size for restaurants that earn less than $50,000 last year.

We need a sample size of n

n is found when $M = 0.04$