3 kg of rump steak costs £42 Adel buys 4 kg of rump steak Work out how much Adel Pays

My cupcake recipe makes $12$ cupcakes and requires $1\frac12$ sticks of butter. I can only buy whole sticks of butter. How many

andre [41]

Answer:

Given that,

My cupcake recipe makes $12$ cupcakes and requires $1\frac12$ sticks of butter. I can only buy whole sticks of butter.

Therefore, 1 whole sticks of butter is enough to make $100$ cupcakes.

8 0

2 years ago

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If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 0

2 years ago

Decide whether the function is an exponential growth or exponential decay function, and find the constant percentage rate of gro

frosja888 [35]

Consider the function f ( x ) = 2479 ⋅ 0.9948x First compare this with f ( x ) po ( 1 + r ) ^ 2 We get po = 2479 And 1 + r = 0.9948 = 1 – 0.0052 r = -0.0052 < 0 Therefore, f is an exponential decay function with a decay rate of 0.0052 x 100 = 0.52%

8 0

2 years ago

The factory quality control department discovers that the conditional probability of making a manufacturing mistake in its preci

Anni [7]

Answer:

The probability that a defective ball bearing was manufactured on a Friday = 0.375

Step-by-step explanation:

Let the event of making a mistake = M

The event of making a precision ball bearing production on Monday = Mo

The event of making a precision ball bearing production on Tuesday = T

The event of making a precision ball bearing production on Wednesday = W

The event of making a precision ball bearing production on Thursday = Th

The event of making a precision ball bearing production on Friday = F

the conditional probability of making a manufacturing mistake in its precision ball bearing production is 4% on Tuesday, P(M|T) = 4% = 0.04

4% on Wednesday, P(M|W) = 0.04

4% on Thursday, P(M|Th) = 0.04

8% on Monday, P(M|Mo) = 0.08

and 12% on Friday = P(M|F) = 0.12

The Company manufactures an equal amount of ball bearings (20 %) on each weekday, Hence, the probability that a random precision ball bearing was made on a particular day of the week, is mostly the same for all the five working days.

P(Mo) = 0.20

P(T) = 0.20

P(W) = 0.20

P(Th) = 0.20

P(F) 0.20

The probability that a defective ball bearing was manufactured on a Friday = P(F|M)

P(F|M) = P(F n M) ÷ P(M)

P(F n M) = P(M n F)

P(M) = P(Mo n M) + P(T n M) + P(W n M) + P(Th n M) + P(F n M)

We can obtain each of these probabilities by using the expression for conditional probability.

P(Mo n M) = P(M|Mo) × P(Mo) = 0.08 × 0.20 = 0.016

P(T n M) = P(M|T) × P(T) = 0.04 × 0.20 = 0.008

P(W n M) = P(M|W) × P(W) = 0.04 × 0.20 = 0.008

P(Th n M) = P(M|Th) × P(Th) = 0.04 × 0.20 = 0.008

P(F n M) = P(M|F) × P(F) = 0.12 × 0.20 = 0.024

P(M) = P(Mo n M) + P(T n M) + P(W n M) + P(Th n M) + P(F n M)

P(M) = 0.016 + 0.008 + 0.008 + 0 008 + 0.024 = 0.064

P(F|M) = P(F n M) ÷ P(M)

P(F n M) = P(M n F) = 0.024

P(M) = 0.064

P(F|M) = P(F n M) ÷ P(M) = (0.024/0.064) = 0.375

Hope this Helps!

4 0

2 years ago

Match the following items.

Margarita [4]

Answer:

$1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40$

Step-by-step explanation:

Given:

$\angle DBC=40$ °

From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.

$m\textrm{ arc CD}=2\times \angle DBC\\m\textrm{ arc CD}=2\times 40=80$

Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,

$m\angle BCD=90$

From the Δ BCD, which is a right angled triangle,

$m\angle DBC+m\angle BDC=90\textrm{ (right angled triangle)}\\40+m\angle BDC=90\\m\angle BDC=90-40=50$

Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.

Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles $m\angle DBC\textrm{ and }m\angle BDC$ and EF is a tangent to the circle at point C.

Therefore, $m\angle DCF=m\angle DBC=40\\m\angle ECB=m\angle BDC=50$

Again, using the same theorem as above,

$m\angle DCF=50\\\therefore m\textrm{ arc BC}=2\times m\angle DCF=2\times 50=100$

Hence, all the angles are as follows:

$1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40$

6 0

2 years ago

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