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4 years ago

Find watch percent increase. round to the nearest percent.

Mathematics

1 answer:

Oksana_A [137]4 years ago

8 0

Hey You!

1. 5 - 8 / 5 × 100 = 60

From $5 to $8 is 60% increase.

2. 20 - 30 / 20 × 100 = 50

From 20 students to 30 students is a 50% increase.

3. 86 - 150 / 86 × 100 = 74.4186046511628 Round to 74.

From 86 books to 150 books is a 74% increase.

4. 3.49 - 3.89 / 3.49 × 100 = 11.461318051575958 Round to 11.

From $3.49 to $3.89 is 11% increase.

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What equation has steeper slope? <br><br> A y = x + 1<br><br> B y = 2x + 6

stich3 [128]

Answer:

its B because

B= -3 and A= -1

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4 years ago

Solve each equation. y/2y+1=2/5

LUCKY_DIMON [66]

Answer: y = i $\sqrt{30}$ /5, -i $\sqrt{30}$ /5

Step-by-step explanation:

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3 years ago

The circle graph shows how Lelia spends her day.

Travka [436]

Answer:

D. 35%

Step-by-step explanation:

Note the percentage of each. You are trying to solve for the combination of homework and school.

In this case,

Homework = 10% of the pie chart.

School = 25% of the pie chart.

Add the two numbers together:

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4 years ago

Read 2 more answers

6. j + 1,239 = 1,300 (1 point) Oj=61 Oj=71 Oj = 59

SSSSS [86.1K]

The answer would be j=61 in order to make the equation true

5 0

3 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago