One of the fastest cars in the world was recently measured covering 5726 meters in 50 seconds.

A hotel charges $20 for the use of its dining room and $2.50 a plate for each dinner. an association gives a dinner and charges

Nina [5.8K]

Answer:

40

Step-by-step explanation:

20 + 2.5x = 3x

20 = 0.5x

x = 40

8 0

1 year ago

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Given that Ray E B bisects ∠CEA, which statements must be true? Select three options. m∠CEA = 90° m∠CEF = m∠CEA + m∠BEF m∠CEB =

FromTheMoon [43]

Answer:

Here is the question attached with.

$m\angle CEA =90 \ (deg)$

$m\angle BEF=135\ (deg)$

$\angle CEF$ is a straight line.

$\angle AEF$ is a right angled triangle.

Options $1,4,5,6$ are correct answers.

Step-by-step explanation:

⇒As $\ ray\ AE$ is ⊥ $FEC$ so it will forms right angled triangle then $m\angle CEA =90\ (deg)$ .

⇒Measure of $\angle BEF =135\ (deg)$ as $\angle BEF =\angle AEB +\angle AEF = (45+90)=135\ (deg)$ as $\angle AEB$ is the bisector of $\angle AEC$ ,meaning that $\angle AEB$ is half of $\angle AEC$ so $\angle AEB = 45\ (deg)$ .

⇒ $\angle CEF$ is a straight line as the angles measure over it is $180\ (deg)$ .

⇒Measure of $\angle AEF = 90\ (deg)$ from linear pair concept.

As $\angle CEA + \angle AEF = 180\ (deg)$ ,plugging the values of $m\angle CEA =90\ (deg)$ we have $\angle AEF = 90\ (deg)$ .

The other two options are false as:

$m\angle CEF=m\angle CEA + m\angle BEF = (90+135)=225$

it is exceeding $180\ (deg)$ whereas $\angle CEF$ is a

straight line.

And $m\angle CEB=2(m\angle CEA)$ is not true.

As $\angle CEA = 90\ (deg)$ and $\angle CEB=45\ (deg)$

So we have total $4$ answers.

The correct options are $1,4,5,6$ .

5 0

2 years ago

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The manager at Gabriela's Furniture Store is trying to figure out how much to charge for a book shelf that just arrived. The boo

BigorU [14]

Answer:

thy should sell it for $235.2 because in the shop they sell everything for 60% the original price for which they bought it

Step-by-step explanation:

4 0

2 years ago

The frequency distribution of weights (in kg) of 40 persons is given below.

gregori [183]

Answer: (a) 4

(b) 5

(c) 14

(d)

Class interval Class mark

30 - 35 $\dfrac{35+30}{2}=32.5$

35 - 40 $\dfrac{35+40}{2}=37.5$

40 - 45 $\dfrac{40+45}{2}=42.5$

45 - 50 $\dfrac{45+50}{2}=47.5$

50 - 55 $\dfrac{50+55}{2}=52.5$

Step-by-step explanation:

The data of 40 persons is given as :

Weights (in kg) Frequency

30 - 35 6

35 - 40 13

40 - 45 14

45 - 50 4

50 - 55 3

(a)

Lower limit is the lowest number in a class interval .

So, is the lower limit of fourth-class interval (45 - 50) is 4.

(b)

Class size = Upper limit - lower limit

So, Class size of first class interval = 35-30=5

Thus , class size of each interval is 5. [ class size remains same]

(c)

Highest frequency in table = 14 which is corresponding to 40 - 45 .

Thus , 40 - 45 is the class interval has the highest frequency.

(d)

Class mark is the mid value of each class interval.

Class interval = $\dfrac{(\text{Upper limit+Lower limit})}{2}$

Class interval Class mark

30 - 35 $\dfrac{35+30}{2}=32.5$

35 - 40 $\dfrac{35+40}{2}=37.5$

40 - 45 $\dfrac{40+45}{2}=42.5$

45 - 50 $\dfrac{45+50}{2}=47.5$

50 - 55 $\dfrac{50+55}{2}=52.5$

8 0

2 years ago

During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim

natita [175]

Answer:

$\frac{2}{9}$

Step-by-step explanation:

Given :

Juan rolled a six-sided number cube 18 times.

The number two occurred four times.

To Find: Juan claimed the experimental probability of rolling a two was approximately 1/9. Why is Juan’s experimental probability incorrect?

Solution:

Total events = number of times cube rolled = 18

Favorable events = The number two occurred four times. = 4

So, Experimental probability = $\frac{\text{Favorable events}}{\text{Total events}}$

= $\frac{4}{18}$

= $\frac{2}{9}$

Thus the experimental probability of rolling a two was $\frac{2}{9}$

So, Juan’s experimental probability was incorrect.

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2 years ago

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