Ask question

Ask question

All categories

3 years ago

6

On a trip to Griffith Observatory, Dave rode his bicycle six more than twice as many miles in the afternoon as in the morning. I

f the entire trip was 57 miles long, then how far did he ride in the morning and in the afternoon?

1 answer:

podryga [215]3 years ago

4 0

Answer:

Dave rode his bicycle 17 miles in the morning

Dave rode his bicycle 40 miles in the afternoon

Step-by-step explanation:

* Lets study the information to solve the question

- Dave rod his bicycle in the morning and again in the afternoon

- Six more than twice as many miles in the afternoon as in the morning

means the distance in the afternoon is 6 more than twice the distance

in the morning

- The entire trip was 57 miles means the total distance in the morning

and in the afternoon was 75

* To solve the question let the distance in the morning is x miles

∵ The distance in the morning = x miles

∵ The distance in the afternoon is 6 more than twice the distance

in the morning

∴ The distance in the afternoon = 2x + 6 miles

∵ The distance in entire trip = 57 miles

- The distance in the morning and the distance in the afternoon = 57 miles

∴ x + (2x + 6) = 57 ⇒ simplify

∴ 3x + 6 = 57 ⇒ subtract 6 from both sides

∴ 3x = 51 ⇒ divide both sides by 3

∴ x = 17 miles

∵ The distance in the morning is x miles

∴ Dave rode his bicycle 17 miles in the morning

∵ The distance in the afternoon = 2x + 6 miles

- Substitute the value of x

∴ The distance in the afternoon = 2(17) + 6 = 34 + 6 = 40 miles

∴ Dave rode his bicycle 40 miles in the afternoon

You might be interested in

Identify the range of the function shown in the graph.

Masteriza [31]

The tan gray tan is black gray or green orange gray green and red violets are gray green and gray and tan and that would be your answer

6 0

2 years ago

What is the arc length of the arc sub tended in a circle with radius 6 and an angle of 7 pi/ 8

KengaRu [80]

Answer:

Formula for the Arc length is given by:

$\text{Arc length} = 2 \pi r \cdot \frac{\theta}{360^{\circ}}$

As per the statement:

radius of circle(r) = 6 units

Angle ( $\theta$ ) = $\frac{7 \pi}{8}$ radian

Use conversion:

$1 rad = \frac{180}{\pi}$

$\frac{7 \pi}{8}$ = $\frac{180}{\pi} \cdot \frac{7 \pi}{8} = \frac{1260}{8} = 157.5^{\circ}$

then;

substitute these given values we have;

Use value of $\pi = 3.14$

$\text{Arc length} = 2\cdot 3.14 \cdot (6) \cdot \frac{157.5^{\circ}}{360^{\circ}}$

or

$\text{Arc length} = 2\cdot 3.14 \cdot (6) \cdot 0.4375$

Simplify:

$\text{Arc length}=16.485$

Therefore, the arc length of the arc substended in a circle with radius 6 units an angle of 7 pi/8 is 16.485 units

7 0

3 years ago

A 2-column table has 3 rows. The first column is labeled Domain with entries 0, 1, 2. The second column is labeled Range with en

skad [1K]

Answer:

Step-by-step explanation:

18 - I took the test on edg.

8 0

3 years ago

Read 2 more answers

Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o

Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is $\hat p_1$ = 1020/1272 = 85/106 = $0.8\overline {0188679245283}$ ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is $\hat p_2$ = 5343/6238 ≈ 0.8565

c) We have the following two way table;

$\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}$

$The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780$

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

$95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}$

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

$95 \% \ CI \approx 0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155$

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0

3 years ago

What is z equal to ? Please help

MakcuM [25]

The sum of the interior angles of a triangle (half the rhombus in this case) is 180. Because of the properties of a rhombus, z and x will be equal to each other. Therefore, $124 + z + z = 180$ .
$124 + z + z = 180$
$124 + 2z = 180$
$2z = 56$
$z = 28$
Thus, z is equal to 28.

7 0

3 years ago