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3 years ago

Can you help me with number 4 and 5 please. I will mark you Brainiest

Mathematics

2 answers:

ra1l [238]3 years ago

6 0

4. Each student gets three donuts and the teacher has 12 remaining for herself.

5. D

Sergio [31]3 years ago

6 0

Answer:

4) 4 donuts are left over 5) D. 21 - 10

Step-by-step explanation:

4) Take the number of donuts, 96, and divide it by the number of students, 28.

96 / 28 = 3 donuts per student with 4 donuts left over.

5) ten less than twenty-one

Less than is a flipper word which means that 21 is subtracted by 10

21 - 10

If these answers are correct, please make me Brainliest!

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An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh

natima [27]

Answer:

Speed of the plane in still air: $779\; {\rm km \cdot h^{-1}}$ .

Windspeed: $100\; {\rm km \cdot h^{-1}}$ .

Step-by-step explanation:

Assume that $x\; {\rm km \cdot h^{-1}}$ is the speed of the plane in still air, and that $y\; {\rm km \cdot h^{-1}}$ is the speed of the wind.

When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be $(x - y)\; {\rm km \cdot h^{-1}}$ .
When this plane is travelling in the same direction as the wind, the ground speed of this plane would be $(x + y)\; {\rm km \cdot h^{-1}}$ .

The question states that when going against the wind ( $v = (x - y)\; {\rm km \cdot h^{-1}}$ ,) the plane travels $6111\; {\rm km}$ in $9\; {\rm h}$ . Hence, $9\, (x - y) = 6111$ .

Similarly, since the plane travels $7911\; {\rm km}$ in $9\; {\rm h}$ when travelling in the same direction as the wind ( $v = (x + y)\; {\rm km \cdot h^{-1}}$ ,) $9\, (x + y) = 7911$ .

Add the two equations to eliminate $y$ . Subtract the second equation from the first to eliminate $x$ . Solve this system of equations for $x$ and $y$ : $x = 779$ and $y = 100$ .

Hence, the speed of this plane in still air would be $779\; {\rm km \cdot h^{-1}}$ , whereas the speed of the wind would be $100\; {\rm km \cdot h^{-1}}$ .

3 0

1 year ago

Sandy's orchard has only 4 apple trees for every 3 pear trees. She is wanting to plant enough avocado trees so that she has 2 av

LUCKY_DIMON [66]

<span>current ratio: 4 apple trees for every 3 pear trees. wants: 2 avocado trees for every 5 pear trees in the orchard. Sandy is planting 18 avocado trees so 2 avocado trees for every 5 pear trees =1 avocado trees for every 2.5 pear trees therefore if we have 18 avocado trees 18 x 2.5 = 45 pear trees 4 apple trees for every 3 pear trees 4/3 = 1.333 apple trees for every 1 pear tree therefore if we have 45 avocado trees 45 x 1.333 = 60 apple trees therefore ratio of avocado to apple trees = 18 : 60 18 : 60 = 9 : 30 = 3 : 10</span>

4 0

3 years ago

Find the value of this expression if x= -7<br><br> x^2+5/x+1

Ray Of Light [21]

Answer:

-9

Step-by-step explanation:

Replace x with -7 in the expression:

x^2 + 5 / x + 1

-7^2 + 5 / -7 + 1

49 + 5 / -6

54/-6

simplify the fraction

9/-1 = -9

8 0

2 years ago

If the cost is $250 each day. How much do I pay at 6 days and 2 nights?

LiRa [457]

Well u see u need to multiply and that's really all to it so yeah ur welcome

6 0

3 years ago

Read 2 more answers

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

NeX [460]

Answer:

(a) 0.343

(b) 0.657

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

$p = \frac{70}{100} = 0.7$

Let P(a vehicle not passing the test) = q

q = 1 - p

q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

= 0.7 × 0.7 × 0.7

= 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

= (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

= 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

= 0.657

= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.063 + 0.063 + 0.063

= 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

= 0.063 + 0.063 + 0.063 + 0.027

= 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

= (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

= 0.973

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

$= \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}$

$= \frac{0.343}{0.973}$

= 0.353

3 0

3 years ago

Read 2 more answers