Find two rational numbers between -1,36 and -1,35

On a trip, Duke encounters some road construction that slows him down. While he is able to go 60 mph on the highway and 40 mph t

natulia [17]

Answer:

Time taken by Duke on the highway = 3 hours

Time spent on the construction = 6 hours

Time spent in the city = 2 hours

Step-by-step explanation:

Speed by which Duke travels a distance on highway $d_1$ = 60 mph

Let the time taken to travel on highway = $t_1$ hours

Distance traveled = $60t_1$ miles

Speed in the city = 40 mph

Let the time taken = $t_2$ hours

Distance traveled in the city = $40t_2$ miles

Speed on the construction = 20 mph

Let the time spend on the construction = $t_3$ hours

Distance traveled on the construction = $20t_3$ miles

Since, total distance covered = 380 miles

$d_1+d_2+d_3=60t_1+40t_2+20t_3$

He spends one half the time on the highway as he does on the construction,

$t_3=2t_1$

Total time spent in the trip = 11 hours

Therefore, $t_1+t_2+t_3=11$

$t_1+t_2+2t_1=11$

$t_2=11-3t_1$

From equation (1),

$60t_1+40(11-3t_1)+20(2t_1)=380$

$60t_1-120t_1+40t_1+440=380$

$440-20t_1=380$

$20t_1=440-380$

$t_1=\frac{60}{20}$

$t_1=3$ hours

$t_2=11-3t_1=2$ hours

$t_3=2t_1=6$ hours

Therefore, time taken by Duke on the highway = 3 hours

Time spent on the construction = 6 hours

Time spent in the city = 2 hours

3 0

3 years ago

The number of texts per day by students in a class is normally distributed with a

kobusy [5.1K]

Answer:

1, 2, 6

Step-by-step explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation$

Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts

1) For x < 90:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2$

From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%

Option 1 is correct

2) For x > 130:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0$

From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%

Option 2 is correct

3) For x > 190:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3$

From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%

Option 3 is incorrect

4) For x < 130:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0$

For x > 100:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5$

From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%

Option 4 is incorrect

5) For x = 130:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0$

Option 5 is incorrect

6) For x = 130:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5$

Since 1.5 is between 1 and 2, option 6 is correct

5 0

2 years ago

Prove the following identity : Sin α . Cos α . Tan α = (1 – Cos α) (1 + Cos α)

Umnica [9.8K]

Let's work on the left side first. And remember that
the tangent is the same as sin/cos.

sin(a) cos(a) tan(a)

Substitute for the tangent:

[ sin(a) cos(a) ] [ sin(a)/cos(a) ]

Cancel the cos(a) from the top and bottom, and you're left with

[ sin(a) ] . . . . . [ sin(a) ] which is [ sin²(a) ] That's the left side.

Now, work on the right side:

[ 1 - cos(a) ] [ 1 + cos(a) ]

Multiply that all out, using FOIL:

[ 1 + cos(a) - cos(a) - cos²(a) ]

= [ 1 - cos²(a) ] That's the right side.

Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.

So, on the right side, you could write [ sin²(a) ] .

Now look back about 9 lines, and compare that to the result we got for the left side .

They look quite similar. In fact, they're identical. And so the identity is proven.

Whew !

4 0

2 years ago

Read 2 more answers

Which of these quadrilaterals have opposite sides that are parallel? Name these quadrilateral

Ad libitum [116K]

Answer:

square,rectangle,rhombus,paralellagram

Step-by-step explanation:

4 0

2 years ago

Write a one-sentence story problem about a ratio.

Serga [27]