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2 years ago

7

Tanvi finds that her robot vacuum can clean her 150-square-foot kitchen in 1 2 of an hour. How many square feet could the robot

vacuum clean in 1 hour?

2 answers:

anastassius [24]2 years ago

8 0

Answer:

Help

Step-by-step explanation:

Julli [10]2 years ago

3 0

Answer:

300

Step-by-step explanation:

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An old rectangular tank with length 0.3 m could hold 154 L of water. A new similar tank has a length of 1.5 m. What is the capac

dimaraw [331]

The length of the new tank is 5 times bigger than the old one, so 154×5=770. As a result, the capacity of the new tank is 770L. Hope it help!

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2 years ago

Read 2 more answers

Owen has enough materials to build up to 10 birdhouses in shop class. Each birdhouse needs 12 square feet of wood. The function

kirill [66]

Answer:

D. D: 0 ≤ b ≤ 10; R: 0 ≤ W(b) ≤ 120

Step-by-step explanation:

The problem statement tells you Owen can build up to 10 birdhouses, and that b represents that number. Then 0 ≤ b ≤ 10 is the domain described by the problem statement.

It also tells you that W(b) = 12b, so filling in values from 0 to 10 gives a range from 0 to 120: 0 ≤ W(b) ≤ 120.

These observations match choice D.

8 0

1 year ago

Given the image below, what is the image of C after a dilation with a scale factor of 3

Ne4ueva [31]

Answer: The coordinates of point C after the dilation are (-2, 5)

Step-by-step explanation:

I guess that you want to find where the point C ends after the dilation.

Ok, if we have a point (x, y) and we do a dilation with a scale A around the point (a,b), then the dilated point will be:

(a + A*(x - a), b + A*(y - b))

In this case we have:

(a,b) = (2,1) and A = 3.

And the coordinates of point C, before being dilated, are: (1, 2)

Then the new location of the point C will be:

C' = (1 + 3*(1 - 2), 2 + 3*(2 - 1)) = (1 -3, 2 + 3) = (-2, 5)

6 0

2 years ago

Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba

Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

<u>For X=0</u>

This happens when every couple and the singles are on time (ot).

$P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602$

<u>For X=1</u>

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

$P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908$

<u>For X=2</u>

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

$P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704$

<u>For X=3</u>

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

$P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055$

<u>For X=4</u>

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

$P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285$

<u>For X=5</u>

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

$P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550$

<u>For X=6</u>

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

$P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427$

<u>For X=7</u>

This happens when

1) Only one of the singles is on time (2 combinations)

$P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390$

<u>For X=8</u>

This happens when everybody is late

$P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147$

8 0

1 year ago

The perimeter of the norwegian flag is 190 inches. What are the dimensions of the flag?

mario62 [17]

<span>The dimensions are 40 inches by 55 inches.

Explanation<span>:
We know that perimeter is the sum of all of the sides. Since this is rectangular, opposite sides are equal. This gives us
y+11/8y+y+11/8y=190.

Combining like terms, we have
2y+22/8y=190.

Writing 22/8 as a mixed number, we have
2y+2 3/4y=190
4 3/4y=190.

Divide both sides by 4 3/4:
(4 3/4y)</span></span>÷<span><span>(4 3/4)=190</span></span>÷<span><span>(4 3/4)
y=190</span></span>÷<span><span>(4 3/4).

Convert the mixed number to an improper fraction:
y=190</span></span>÷<span><span>(19/4).

To divide fractions, flip the second one and multiply:
y=190*(4/19)=760/19=40.

Since y=40, 11/8y=11/8(40)=440/8=55.</span></span>

8 0

2 years ago