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1 year ago

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Hiro has a stack of cards with one number from the set 1, 1, 2, 2, 3, 3, 3, 4 written on each card. What is the probability that

he pulls out a 3 first and then pulls out a 2 without replacing them?
a. 1/64
b. 1/56
c. 3/32
d. 3/28

1 answer:

professor190 [17]1 year ago

7 0

Answer:

Therefore, the probability is P=3/32.

Step-by-step explanation:

We know that Hiro has a stack of cards with one number from the set 1, 1, 2, 2, 3, 3, 3, 4 written on each card.

We calculate the probability that he pulls out a 3 first and then pulls out a 2 without replacing them.

The probability that he pulls out a 3 first is 3/8.

The probability of a second card being 2 is 2/8.

We get:

$P=\frac{3}{8}\cdot \frac{2}{8}\\\\P=\frac{6}{64}\\\\P=\frac{3}{32}$

Therefore, the probability is P=3/32.

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Amar cycles at a speed of 18km/h.

damaskus [11]

Answer:

<em>The distance between the two villages is 16.5 Km</em>

Step-by-step explanation:

<u>Constant Speed Motion</u>

It's a type of motion in which the distance of an object changes by an equal amount in every equal period of time.

If v is the constant speed, the object travels a distance x in a time t, given by the equation:

x=vt

Amar cycles at v=18 Km/h for t=55 minutes. We need to calculate the distance traveled between the two villages.

Since the speed and the time are given in different units, we convert the time to hours, recalling that 1 hour=60 minute.

t=55 min = 55/60 hours

For the sake of precision, we won't operate the division so far. Compute the distance:

x=18 *55/60=16.5 Km

The distance between the two villages is 16.5 Km

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2 years ago

during the summer winnie delivers flowers and bob babysits for neighbors to show their earnings winnie makes a table which state

miv72 [106K]

Winnie earns $12.30 and Bob earns $24.30.
Bob earns $8.10 and earns$4.10
Winnie earns$12.15 and Bob earns $6.15
Winnie earns $4.05 and Bob earns $2.05

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2 years ago

Read 2 more answers

Please answer all of them need this

VikaD [51]

First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in $\Delta BCQ$ , we can easily find the values of CQ and BQ.

Since, $Sin(75^0)=\frac{CQ}{8}$

$\therefore CQ=8\times Sin(75^0)\approx 7.73$ ft

In a similar manner we can find BQ as:

$Cos(75^0)=\frac{BQ}{8}$

$BQ\approx2.07$ ft

All these values can be found in the diagram attached.

Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:

$PQ=22-(AP+QB)=22-(2.07+2.07)=17.86$

Let us now consider $\Delta AQC$

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of $\Delta AQC$ .

$AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38$ feet.

Thus, out of the given options, Option B is the closest and hence is the answer.

Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area= $\frac{1}{2}$ (First Side)(Second Side)(Sine of the angle between the two sides)

Thus, from the given data,

Area= $\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845$ $m^2$

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the $\Delta ABC$ given to us.

Thus, from the triangle we will have:

$\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}$

$\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}$

$\frac{17}{Sin(25^0)}=\frac{a}{Sin(45^0)}$

This gives $a$ to be:

$a\approx28.44$

Which is not close to any of the given options.

Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

$AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}$

$\therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8$ miles.

Thus, Option D is the answer.

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2 years ago

Read 2 more answers

The heights of the trees for sale at two nurseries are shown below. Heights of trees at Yard Works in feet : 7, 9, 7, 12, 5 Heig

Troyanec [42]

Answer:

The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

Step-by-step explanation:

1). Height of the trees at Yard Works are = 7,9,7,12,5 feet

So mean height of the trees = (7+9+7+12+5)÷5

= 40÷5 =8 feet

Standard deviation of the trees at Yard works = ∑(║(height of the tree-mean height of the tree))║/(number of trees)

(height of the tree-mean height of the tree)= ║(7-8)║+║(9-8)║+║(7-8)║+║(12-8)║+║(5-8)║ = (1)+1+(1)+4+(3)= 10

Therefore standard deviation = (10)/(5) =2

2). In the same way mean height of the trees at Grow Station=(9+11+6+12+7)/5= 45/5 = 9

Now we will calculate the mean deviation of the tress at Grow Station

= ∑║(height of the tree-mean height of the tree)║/(number of trees)

= ║(9-9)║+║(11-9)║+║(6-9)║+║(12-9)║+║(7-9)║/(5)

= (0+2+3+3+2)/5

= 10/5 =2

Therefore The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

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2 years ago

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Customer: "I buy the same items every week and my bill is normally $90.00. Why am I now being charged $108.00 for the exact same

scZoUnD [109]

18
$108.00- 90.00= 18.00$

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2 years ago