Which statement is NOT true?

A bus with kids and a truck with their bags started moving from the school to the camp at the same time. The speed of the bus wa

natta225 [31]

Answer:

1 hour

Step-by-step explanation:

The distance covered by each of the two vehicle can be written as

$d=vt$

where

d is the distance covered

v is the speed of the vehicle

t is the time elapsed

For the bus, the speed is

$v_b = 60 mph$

So the distance covered by the bus in a time t is

$d=60t$ (1)

The kids in the bus took 3 hours to reach the camp,

t = 3 h

So the distance between the school and the camp is:

$d=(60)(3)=180 mi$

For the truck, the speed is

$v_t=45 mph$

So the distance covered by the truck in a time t is

$d=45t$ (2)

Since the distance to cover is 180 miles, the time taken by the truck is

$t=\frac{d}{45}=\frac{180}{45}=4 h$

And so, the truck arrives

$\Delta t = t_t - t_b = 4 h - 3 h = 1 h$

1 hour after the bus.

3 0

4 years ago

Mai has picked 1 cup of strawberries for a cake, which is enough for ¾ of the cake. How many cups does she need for the whole ca

Kitty [74]

Answer:

1 1/4 cups

Step-by-step explanation:

1 cup= 3/4 of a cake

1 1/4= 1 whole cake

I hope this is what you're looking for.

8 0

3 years ago

The opponents of soccer team A are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores

NikAS [45]

Answer:

a) The expected number of goals team A will score is 5.1

b) The probability that team A will score a total of 5 goals is 0.1147

Step-by-step explanation:

Let X be the amount of goals scored by team A in both matches. Let X1 and X2 be the total amount of goals team A scores in match 1 and 2 respectively, then X = X1+X2, and also

a)

E(X) = E(X1+X2) = E(X1)+E(X2) = 0.6*2+0.4*3 + 0.3*2+0.7*3 = 5.1

b) In order for X to be equal to 5 we have 5 possibilities

- X1 is 0 and X2 is 5

- X1 is 1 and X2 is 4

- X1 is 2 and X2 is 3

- X1 is 3 and X2 is 2

- X1 is 4 and X2 is 1

- X1 is 5 and X2 is 0

Let T1 be a poisson distribution with mean λ = 2, then

$P(T1=0) = e^{-2}$

$P(T1=1) = 2 * e^{-2}$

$P(T1=2) = 2 * e^{-2}$

$P(T1=3) = \frac{4}{3} \, e^{-2}$

$P(T1=4) = \frac{2}{3}\, e^{-2}$

$P(T1=5) = \frac{4}{15}\,e^{-2}$

Lets do the same with a Poisson distribution T2 with mean λ = 3

$P(T2=0) = e^{-3}$

$P(T2=1) = 3 \, e^{-3}\\P(T2=2) = \frac{9}{2} \, e^{-3}\\P(T2=3) = \frac{9}{2} \, e^{-3}\\P(T2=4) = \frac{27}{8} \, e^{-3}\\P(T2=5) = \frac{81}{40} \, e^{-3}$

Now, we are ready to compute the probability that X is equal to 5.

$P(X1 = 0, X2 = 5) = (0.6* e^{-2} + 0.4*e^{-3}) * (0.3*\frac{4}{15}e^{-2} + 0.7*\frac{81}{40} e^{-3}) = 0.00823\\P(X1 = 1, X2 = 4) = (0.6* 2e^{-2} + 0.4*3 e^{-3}) * (0.3*\frac{2}{3}e^{-2} + 0.7*\frac{27}{8} e^{-3}) = 0.03214\\P(X1 = 2, X2 = 3) = (0.6* 2e^{-2} + 0.4*\frac{9}{2}e^{-3}) * (0.3*\frac{4}{3}e^{-2} + 0.7*\frac{9}{2} e^{-3}) = 0.05317\\P(X1 = 3, X2 = 2) = (0.6* \frac{4}{3}e^{-2} + 0.4*\frac{9}{2}*e^{-3}) * (0.3*2e^{-2} + 0.7*\frac{9}{2} e^{-3}) = 0.0471$

$P(X1 = 4, X2 = 1) = (0.6* \frac{2}{3}e^{-2} + 0.4*\frac{27}{8}e^{-3}) * (0.3*2e^{-2} + 0.7*3e^{-3}) = 0.0225\\P(X1 = 5, X2 = 0) = (0.6* \frac{4}{15}e^{-2} + 0.4*\frac{81}{40}e^{-3}) * (0.3*e^{-2} + 0.7*e^{-3}) = 0.0047$

We can conclude that

P(X = 5) = 0.00823+0.03214+0.05317+0.0471+0.0225+0.0047 = 0.1147

The probability that team A will score a total of 5 goals is 0.1147

7 0

3 years ago

Could someone give me a list of the common factors? The numbers are 150 and 135. Thanks :)

dexar [7]

First, let's list the factors of each number
150: 1, 150, 2, 75, 3, 50, 5, 30, 6, 25, 10, 15
135: 1, 135, 3, 45, 5, 27, 9, 15
From these lists, we can say for certain that 1, 3, 5, and 15 go into both numbers, so that would be your list of the common factors.

I hope this helps!

4 0

3 years ago

Pleas help ill give brainiest if your answers correct ok pleas list them 1-5

Sindrei [870]

Answer:

Descending order

Step-by-step explanation:

$1. \sqrt{25} = 5\\2. \frac{5}{2} = 2.5\\3. 2.51\times10^{-1} = 0.251 \\4. 0.25 = 0.25\\5. -2.5\% = -0.025$

6 0

3 years ago