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3 years ago

On June 1, a fast-growing species of algae is accidentally introduced into a lake in a city park. It starts to grow and

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

4 0

Answer:

June 29

Step-by-step explanation:

Since the question clearly states that the rate of algae growth is such that the area it covers doubles every day then the inverse is also true, i.e. going back a day the area covered is half.

It can be simply assumed that on the previous date from the day the entire lake was covered, half of the area must have been covered. This means that on the day prior to June 30 the lake was half covered, i.e. June 29

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Mrs. Travis wants to have a clown deliver balloons to her secretary's office. Clowns R Fun charges $ 1.25 per balloon and $ 6 de

choli [55]

Let C be Clowns R Fun charge for b balloons and S be Singing Balloons charge.
C=6+1.25b and S=2+1.95b. C must be less than S.
6+1.25b<2+1.95b, 4<0.7b, b>4/0.7, b>40/7, b>5. So Mrs Travis needs to purchase more than 5 balloons.

6 0

3 years ago

ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions

ololo11 [35]

Answer:

The answer would be f(a)=-1

-AC I hope this helps<3

7 0

3 years ago

A miniature quadcopter is located at xi = −1.75 m and yi = −5.70 m at t = 0 and moves with an average velocity having components

Black_prince [1.1K]

The quadcopter's average velocity is given in terms of change in position by

$\vec v_{\rm av}=\dfrac{\Delta x}{\Delta t}\,\vec\imath+\dfrac{\Delta y}{\Delta t}\,\vec\jmath$

where $\Delta x=x_f-x_i$ , the difference in the quadcopter's final and initial positions and $\Delta t=(1.60-0)\,\mathrm s=1.60\,\mathrm s$ .

The $x$ -component of the average velocity is 2.70 m/s, so

$2.70\dfrac{\rm m}{\rm s}=\dfrac{x_f-(-1.75\,\mathrm m)}{1.60\,\rm s}\implies \boxed{x_f=2.57\,\mathrm m}$

and the $y$ -component is -2.50 m/s, so

$-2.50\dfrac{\rm m}{\rm s}=\dfrac{y_f-(-5.70\,\mathrm m)}{1.60\,\rm s}\implies\boxed{y_f=-9.70\,\mathrm m}$

4 0

3 years ago

There are 412 students and 20 teachers taking buses on a trip to a museum. Each bus can seat a maximum of 48. Which inequality g

-BARSIC- [3]

The inequality $\mathrm{b} \geq \frac{\text { total number of travelers }}{\text { capacity of each bus }}$ gives the least number of buses, b, needed for the trip. The least number of buses is 9

Solution:

Given that, There are 412 students and 20 teachers taking buses on a trip to a museum.

Each bus can seat a maximum of 48.

We have to find which inequality gives the least number of buses, b, needed for the trip?

Now, there are 412 students and 20 teachers, so in total there are 412 + 20 = 432 travelers

The number of buses required “b” is given as:

$\text { (b) } \geq \frac{\text { total number of travelers }}{\text { capacity of each bus }}$

$\text { So, number of buses required } \geq \frac{432}{48}$

Number of buses required ≥ 9 buses.

But least number will be 9 from the above inequality.

Hence, the inequality $\mathrm{b} \geq \frac{\text {total number of travelers}}{\text {capacity of each bus}}$ gives least count of busses and least count is 9.

6 0

3 years ago

The local branch of the Internal Revenue Service spent an average of 21 minutes helping each of 10 people prepare their tax retu

AURORKA [14]

Answer:

The confidence Interval is [- 0.7053 10.4521]

a: The hypotheses are

H0: μ1=μ2 against the claim Ha :μ1≠μ2

b. The critical value for t∝/2 for 17 d.f t > 2.508 and t < -2.111

c. t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

Step-by-step explanation:

When the standard deviations are not the same then the confidence intervals for mean differences are calculated as

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

x1`= 21 x2`= 27

n1= 10 n2= 14

s1= 5.6 s2= 4.3

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

The t∝/2 for 17 d.f = 2.11

Putting the values

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

(21-27) - 2.11√5.6²/10+ 4.3²/14 < u1-u2 <(21-27) +2.11√5.6²/10+4.3²/14

6- 2.11*2.111 < u1-u2 < ( 6 ) +2.11*2.111

6- 4.4521 < u1-u2 < ( 6 ) +5.294

- 1.5479 < u1-u2 < 10.4521

The confidence Interval is [- 0.7053 10.4521]

a: The hypotheses are

H0: μ1=μ2 against the claim Ha :μ1≠μ2

The claim is that there is a difference in the average time spent by the two services

b. The critical value for t∝/2 for 17 d.f t > 2.508 and t < -2.111

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

c. The test statistic is

t= (x1`-x2`) /√s1²/n1 + s2²/n2

t= (21-27) /√5.6²/10+ 4.3²/14

t= -6/2.111

t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

5 0

3 years ago