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3 years ago

Solve for b b/2 - 5 =4 with steps

Mathematics

2 answers:

damaskus [11]3 years ago

7 0

Answer:

Step-by-step explanation:

b/2-5=4 ADD 5 TO BOTH SIDES

+5 +5

b/2=9

b/2=9/1 CROSS MULTIPLY

b=18

jok3333 [9.3K]3 years ago

6 0

Answer:

b = 18

Step-by-step explanation:

b/2 - 5 = 4

b/2 = 4 + 5

b/2 = 9

b = 9 x 2

b = 18

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I NEED HELP WITH NUMBER FIVE- NUMBER FOURS ANSWER IS 108

Nookie1986 [14]

Answer:

s = 36b

Step-by-step explanation:

252/7 = 36, so maximum of 36 students can fit in one bus. The equation is s = 36b.

6 0

2 years ago

Read 2 more answers

The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt

AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average height = 175 cm

$\sigma$ = population standard deviation = 20 cm

n = sample of men = 40

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $175-1.96 \times }{\frac{20}{\sqrt{40} } }$ , $175+1.96 \times }{\frac{20}{\sqrt{40} } }$ ]

= [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.

3 0

3 years ago

Circles and segments create the many relationships addressed in the theorems throughout this unit. Use those theorems and the in

ohaa [14]

In your questions where the segment and circles create many relationship addressed in the theorem throughout this unit. So the following are the answers:
b.KI=5
c. EP = 2
e. CJ = 6
f.FH=14.4
g.AE=8.75

I hope you are satisfied with my answer

8 0

3 years ago

In a particular class, exams are worth 50% of the overall grade, quizzes are worth 30% of the overall grade, homework is worth 1

UkoKoshka [18]

Answer:

His overall grade is 81.9%

Explanation:

The easiest way to solve this question is by using simple cross multiplication.

1- His overall percentage in exams:

We are given that exams are worth 50% of the grade and that he scored an average of 82.2%

Assume that his overall percentage in exams is e

We can setup the ratio and solve for e as follows:

$\frac{82.2}{100}=\frac{e}{50} \\ \\e=\frac{50*82.2}{100}=41.1$ %

2- His overall grade in quizzes:

We are given that quizzes are worth 30% of the grade and that he scored 74.4%

Assume that his overall percentage in quizzes is q

We can setup the ratio and solve for q as follows:

$\frac{74.4}{100}=\frac{q}{30} \\ \\q=\frac{30*74.4}{100}=22.32$ %

3- His overall grade in homework:

We are given that homework is worth 15% of the grade and that he scored an average of 90%

Assume that his overall percentage in exams is h

We can setup the ratio and solve for h as follows:

$\frac{90}{100}=\frac{h}{15} \\ \\h=\frac{15*90}{100}=13.5$ %

4- His overall grade in class participation:

We are given that class participation is worth 5% of the grade and that he scored a 100% in it

Assume that his overall percentage in exams is p

Since he scored 100%, then he scored the complete worth which is 5%

p = 5%

5- His overall grade in the course:

The student's overall grade in the course is the summation of the above calculated percentages

This means that:

Overall grade = e + q + h + p

Overall grade = 41.1 + 22.32 + 13.5 + 5

Overall grade = 81.92% which is 81.9% approximated to one decimal place

Hope this helps :)

3 0

3 years ago

Two trains leave the station at the same time, one heading west and the other east. The westbound train travels 14 miles per hou

Sergeu [11.5K]

Answer:

The answer to your question is 166 mi/h

Step-by-step explanation:

Data

ve = rate eastbound train

vw = rate westbound train = ve - 14

distance = d = 900 mi

time = t = 5 h

Formula

v = $\frac{distance}{time} = \frac{d}{t}$

Substitution and simplification

$ve = \frac{900}{5} = 180 mi/h$

Now, substitute this value in the equation of vw

vw = 180 - 14

Result

vw = 166 mi/h

7 0

3 years ago