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2 years ago

If you spend 780 minutes per year waiting at traffic lights, how many hours is this?

Mathematics

2 answers:

rjkz [21]2 years ago

8 0

Answer:

13 hours

Step-by-step explanation:

minutes spent waiting on light/60= hours spent waiting on light

Tanzania [10]2 years ago

3 0

Answer:
That will me 13 hours

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Dev has 9 shells. Zoe has 55 shells, Zoe gives some shells to Dev. Now zoe has 3 times as many shells as dev. How many shells do

Goryan [66]

Answer:

number of shells zoe gives to dev = 7

number of shells zoe is left with = 55-7= 48

number of shells dev has = 9+7=16

Step-by-step explanation:

let the initial number of shells that dev has be A, and initial number of shells that zoe has be B.

let the number of shells that zoe gives to dev be x.

after giiving x shells zoe is left with 3 times the number of shell as that of dev.

therefore number of shells with zoe = 3×number of shells with dev.

number of shells with zoe = initial shells - x = 55-x

number of shells with dev = initial shells + number of shells he gets

= 9+x

therefore (55-x)=3×(9+x)

55-x = 27+3x

55-27=3x+x

4x = 28

x= 7

therefore number of shells zoe gives to dev = 7

number of shells zoe is left with = 55-7= 48

number of shells dev has = 9+7=16

6 0

2 years ago

M<2=34 find m<4 and explain how you know

Naily [24]

Angle 4 equals 34 degrees since angle 2 and 4 are opposite angles and opposite angles are congruent

6 0

2 years ago

The percent decrease from 12 to 9 is equal to the percent decrease from 40 to what number

Zina [86]

(12 - 9) / 12 = (40 - x) / 40
3/12 = (40 - x) / 40
40(1/4) = 40 - x
10 = 40 - x
10 - 40 = -x
-30 = -x
30 = x <=== percent decrease from 40 to 30

8 0

1 year ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

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

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 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago

An investment earns 6% simple interest. What amount of principal is needed to have $3000 after 5 years? Round your answer to the

amid [387]

I = p*r*t
$3000=p*.06*5=?
Multiply and you would have yor answer(rounded to nearsest cent)!

6 0

2 years ago

Read 2 more answers