A man travels from Town X to Town Y at an average rate of 60 mph and returns at an average rate of 50 mph. He takes a 1/2 hour longer than he would take if he made the round trip at an average of 55 mph. What is the distance from Town X to Town Y?

Answers

Answer 1

Answer: $v_1=60mph\n v_2=50mph\n v_3=55mph\n t_1- time\ needed\ to\ travel\ from\ X\ to\ Y\ at\ v_1\ and\ return\ with\ v_2\n t_2- time\ needed\ to\ travel\ from\ X\ to\ Y\ at\ v_3\n\n t_1-0,5h=t_2\n s-travelled\ distance\n v=(s)/(t)\n v*t=s\n t=(s)/(v)\n (0,5s)/(60)+(0,5s)/(50)-0,5=(s)/(55)\n\n (1)/(120)s+(1)/(100)s-(1)/(55)s=(1)/(2)\n (1)/(6600)s=(1)/(2)\n s=3300miles\n Distance\ from\ town\ X\ to\ town\ Y\ is 1650miles.$

Answer 2

Answer:

Answer:

1650 miles

Step-by-step explanation:

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Olivia ordered 24 cupcakes and a layer cake. The layer cake cost $16, and the cost of the order was $52. What was the price of each cupcake?

An photo studio is going to display a portrait in their lobby. The portrait is being reproduced from a photo with dimensions 6 inches by 3 inches. If the dimensions of the portrait in the lobby are 20 times the dimensions of the original photo, determine the area of the portrait in square feet.a.
7,200 ft2

b.
50 ft

c.
300 ft

d.
200 ft

Answers

Answer:

300ft

Step-by-step explanation:

The area of a rectangular city park is 25/54 square miles. The length of the park is 5/9 mile. what is the width in miles of the park?

Answers

$(25)/(54)$ divided by $(5)/(9)$
= $(25)/(54)$ multiplied by $(9)/(5)$
= $(5)/(6)$ miles

Roseanna did not pay last month's credit card bill in full. Her daily balances for her last billing cycle are listed. What was Roseanna's average daily balance to the nearest cent? (Hint: Find the average or mean of these numbers.)For seven days she owed $456.11

For three days, she owed $1,077.00

For six days, she owed $900.08

For nine days, she owed $2,113.15

For five days, she owed $2,310.15

Answers

Average daily balance = (456.11 + 1,077.00 + 900.08 + 2,113.15 + 2,310.15)/(7 + 3 + 6 + 9 + 5) = 6,856.49/30 = 228.55

Therefore, Roseanna's average daily balance = $228.55

Answer:

$228.54

Step-by-step explanation:

Given : Roseanna did not pay last month's credit card bill in full. Her daily balances for her last billing cycle are listed.

To Find: What was Roseanna's average daily balance to the nearest cent?

Solution:

For seven days she owed $456.11

For three days, she owed $1,077.00

For six days, she owed $900.08

For nine days, she owed $2,113.15

For five days, she owed $2,310.15

Mean = $\frac{\text{Sum of amount owned}}{\text{Total no. of days}}$

Mean = $(456.11+1077+900.08+2113.15+2310.15)/(7+3+6+9+5)$

Mean = $(6856.49)/(30)$

Mean = $228.5496$

Hence Roseanna's average daily balance to the nearest cent is $228.54

A fisherman can row upstream at 4mph and downstream at 6mph. He started rowing upstream until he got tired and then started downstream to his start point. How far did the fisherman row if the trip took 3 hours?

Answers

Answer:

Fisherman traveled 14.4 miles.

Step-by-step explanation:

A fisherman's downstream speed = 6 mph

and upstream speed = 4 mph

Let the fisherman traveled x miles he travels upstream.

Time spent in travelling x miles = $(x)/(4)$ hours [Since time = $\frac{\text{distance}}{\text{speed}}$ ]

Similarly when fisherman covers x miles down stream

Time spent in x miles = $(x)/(6)$ hours

Now total time in going upstream and downstream is 3 hours then

$(x)/(4)+(x)/(6)=3$

x $((3+2)/(12))=3$

$((5x)/(12))=3$

5x = 12×3

5x = 36

x = $(36)/(5)=7.2$ miles.

Since fisherman has traveled 2x distance so the answer will be 14.4 miles

Therefore, fisherman traveled 14.4 miles.

total time = upstream time + downstream time
he rowed the same distance both ways.
upstream time = D/4
downstream time = D/6
total time = D/4 + D/6 = 3D/12 + 2D/12 = 5D/12 = 3 hours
D = 3*12/5 = 36/5 or 7.2 miles
where D is the one-way distance. The total distance he rowed is 2*D = 72/5 or 14.4 miles

In the circle, m arc BC = 72°. The diagram is not drawn to scale.

What is m angle BCP

Answers

Answer:

Measure of ∠BCP is 36°

Step-by-step explanation:

Given the circle in which measure of arc BC is 72°

we have to find the measure of angle BCP.

m∠BOC=∠2=72°

By theorem angle subtended at the centre is twice the angle formed at the circumference of circle.

∴ ∠2=2∠1 ⇒ 72=2∠1

⇒ ∠1=36°

By alternate segment theorem which states that

The angle formed between a chord and a tangent through one of the end points of chord is equals to angle in alternate segment.

⇒ ∠3=∠1=36°

Hence, measure of ∠BCP is 36°

Option D is correct

36 you take 72 and divide it by 2 and get 36

Deanna collected data on the favorite sports of the students of two grades. The table shows the relative frequencies of rows for the data collected:Favorite Sport
Swimming Running Volleyball Row totals
Grade 8 0.14 0.18 0.22 0.54
Grade 9 0.17 0.24 0.05 0.46
Column totals 0.31 0.42 0.27 1

Based on the data, which statement is most likely correct?
In Grade 8, 14 students liked swimming.
In Grade 9, 31% of students liked swimming.
In Grade 8, 22% of students liked volleyball.
In Grade 9, 5 students liked volleyball.

Answers

Answer:

In Grade 8, 22% of students liked volleyball.

Step-by-step explanation:

Given:

The table representing the relative frequencies of different sports in different grades among the students.

Swimming Running Volleyball Row totals

Grade 8 0.14 0.18 0.22 0.54

Grade 9 0.17 0.24 0.05 0.46

Column totals 0.31 0.42 0.27 1

Relative frequency givesthe ratio of the number of quantities of a particular kind and the total number of quantities.

$\textrm{Relative frequency}=\frac{\textrm{Number of students in a particular sport}}{\textrm{Total students}}$

From the above table, we can conclude the following points:

Grade 8:

14% liked swimming, 18% liked running and 22% liked volleyball. So, a total of 54% students liked playing sports.

Grade 9:

17% liked swimming, 24% liked running and 5% liked volleyball. So, a total of 46% students liked playing sports.

Combining students of grade 8 and 9:

31% liked swimming, 42% liked running and 27% liked volleyball.

From all the options available, we observe that, third option is only correct.

In Grade 8, 22% of students liked volleyball.

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