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son4ous [18]
3 years ago
6

Help me plzzzzzzzzzzzzzzzzzz.

Mathematics
1 answer:
poizon [28]3 years ago
6 0

Answer:

<u>Option A </u>is your answer

Step-by-step explanation:

hope it helps you

have a great day

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85% of the people survey thought the price of the car wash was reasonable. If 164 people thought the cost of the car wash was re
8090 [49]

192 people were survey

Step-by-step explanation:

The given is:

  • 85% of the people survey thought the price of the car wash was reasonable
  • 164 people thought the cost of the car wash was reasonable

We need to find how many people were survey

Assume that x people were survey

∵ There were x people survey

∵ 85% of them thought the price of the car wash was reasonable

∴ The number of the people who thought the price of the car wash

   was reasonable = 85% × x

∵ 85% = \frac{85}{100} = 0.85

∴ The number of the people who thought the price of the car wash

   was reasonable = 0.85 x

∵ 164 people thought the cost of the car wash was reasonable

- Equate 0.85 x and 164

∴ 0.85 x = 164

- Divide both sides by 0.85

∴ x = 192.94

∵ x represents a number of people, then it must be integer

∴ The number of people were survey = 192

192 people were survey

Learn more:

You can learn more about percentage in brainly.com/question/12960754

#LearnwithBrainly

7 0
3 years ago
Examine the following three premises: i. All kids who have a smartphone play mobile legends.
atroni [7]

Answer:

The increasing number of students who are hooked on playing online mobile games (OMG) is alarming. As such, this study was realized to address the problem. This study assessed the gaming profile towards OMG and its relation to the academic performance of the engineering students of Eastern Visayas State University Tanauan Campus (EVSUTC). Specifically, the study investigated the correlation between student's number of hours spent on playing OMG (at school and home), commonly played OMG (at school and home), reasons for playing OMG and attitudes on playing OMG with academic performance utilizing Eta and Pearson r correlation analyses. A random sample of 134 student respondents were selected through purposive sampling of those who are playing OMG using their mobile phones. Descriptive correlational research design was utilized and a validated survey instrument was employed to gather the needed information. The findings revealed that majority of the students played mobile legends and spent mostly 2 hours playing OMG for a reason of boredom. The overall attitudes of the students on playing OMG were interpreted as Less Favorable (M=2.58, SD=1.13). Out of the independent variables being set in the study, the number of hours spent on playing OMG at home (r=-0.188, p=0.039) and commonly played OMG at school (r=0.203, p=0.045) were found significantly correlated with student's academic performance. Hence, the students' time spent on playing OMG at home and the type of games that students played at school have significant bearing to their academic performance. As such, delimiting student's usage of internet can be made to address the problem.

3 0
2 years ago
Which equation can be used to find the measure of Arc E H G?
anygoal [31]

Answer:

I need more context ,like the graph to figure this out.

Step-by-step explanation:

5 0
3 years ago
Do you know What is X=9+y-2
Damm [24]

Answer:

Subtract

2

from

9

.

X

=

y

+

7

Step-by-step explanation:

5 0
3 years ago
Use the normal distribution of SAT critical reading scores for which the mean is 502 and the standard deviation is 116. Assume t
PSYCHO15rus [73]

Answer:

(a) 93.19%

(b) 267.3

Step-by-step explanation:

The population mean and standard deviation are given as 502 and 116 respectively.

Consider, <em>X</em> be the random variable that shows the SAT critical reading score is normally distributed.

(a) The percent of the SAT verbal scores are less than 675 can be calculated as:

P(X

Thus, the required percentage is 93.19%

(b)

The number of SAT verbal scores that are expected to be greater than 575 can be calculated as:

P(X>575)=P(\frac{x-502}{116}>\frac{575-502}{116}\\P(X>575)=P(Z>\frac{575-502}{116})\\P(X>575)=P(Z>0.6293)\\P(X>575)=0.2673

So,

Out of 1000 randomly selected SAT verbal scores, 1000(0.2673) = 267.3 are expected to have greater than 575.

8 0
3 years ago
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