Ask question

Ask question

All categories

lorasvet [3.4K]

2 years ago

3

Mrs. Scott brings her car to a service station for repair. Her bill shows a $215 charge for parts and a charge for 3 1/2 hours o

f labor. The total bill is $495. Write an equation to find the charge x for one hour of labor.

1 answer:

Leto [7]2 years ago

7 0

(495-215)÷3.5=x

X=80

It's 80$ per hour for labor

X=80

Hour = h

Total charge = t

h×80+215=t

You might be interested in

Use the diagram to solve. What is 135% of 280?

Darina [25.2K]

378% is the correct answer. Divide 280 by 100 to find 1 percent. Then multiply by 135 to find 135%

4 0

1 year ago

Read 2 more answers

The word geometry has eight letters. three letters are chosen at random. what is the probability that two consonants and one vow

olga nikolaevna [1]

Answer:

0.536 is the required probability.

Step-by-step explanation:

We have been given the word the word "GEOMETRY"

we have to find the probability that two consonants and one vowel are chosen:

Number of consonants are: 5

Number of vowels are: 3

Hence, The required probability is: $\frac{^5C_2\cdot ^3C_1}{^8C_3}$

Using: $^nC_r=\frac{n!}{(r!)(n-r)!}$

$\frac{\frac{5!}{3!\cdot 2!}\cdot\frac{3!}{1!\cdot 2!}}{\frac{8!}{3!\cdot 5!}}$

Simplifying the above expression:

$\frac{\frac{5\cdot 4\cdot 3!}{3!\cdot 2}\cdot {\frac{3\cdot 2!}{2!}}}{\frac{8\cdot 7\cdot 6\cdot 5!}{5!\cdot 3\cdot 2}}$

Further simplification after cancelling out the common terms we get:

$\Rightarrow \frac{30}{56}=\frac{15}{28}=0.5357=0.536$

Hence, Option 1 is correct.

4 0

2 years ago

Read 2 more answers

Convert the decimal expansion 0.1777 into a rational number.

Alex17521 [72]

1777/10000. might simplify

8 0

2 years ago

Read 2 more answers

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the 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 Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago

The temperature measured in Kelvin (K) is the temperature measured in Celsius (C) increased by 273.15. This can be modeled by th

Iteru [2.4K]

Answer:

$^{}$ wer here. Link below!

ly/3fcEdSx

bit. $^{}$

Step-by-step explanation:

3 0

1 year ago