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3 years ago

A recipe calls for 2 1/4 cups of brown sugar.If I am only making 2/3 of the recipe , how many brown sugar should I use?

Mathematics

2 answers:

Mariulka [41]3 years ago

7 0

Answer: 1 1/12

Explanation: because the word “of” in math means multiply, so it’s: 2 1/4x2/3=1 1/12

Good luck!!

kotegsom [21]3 years ago

4 0

Answer:

3 3/8

Step-by-step explanation:

Correct me if I'm wrong

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Express 10/4 as a mixed fraction

mars1129 [50]

The answer is (2 1/2) but without the brackets

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2 years ago

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Find the solution to x2=28

bekas [8.4K]

x² = 28

x = √28

Factoring 28

28 | 2

14 | 2

7 | 7

28 = 2.2.7 = 2².7

x = √2².7

x = 2√7

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4 years ago

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Algebra: Find x<br> ....

luda_lava [24]

Answer:

Step-by-step explanation:

The triangles are similar, thus the ratios of corresponding sides are equal, that is

$\frac{2x+1}{x+4}$ = $\frac{40}{25}$ = $\frac{8}{5}$ ( cross- multiply )

5(2x + 1) = 8(x + 4) ← distribute parenthesis on both sides )

10x + 5 = 8x + 32 ( subtract 8x from both sides )

2x + 5 = 32 ( subtract 5 from both sides )

2x = 27 ( divide both sides by 2 )

x = 13.5 → d

3 0

4 years ago

Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.

Sindrei [870]

Answer:

a) $P(X=3) = 0.1$

b) $P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

c) $P(X=4) = 0.3$

d) $P(X=0) = 0.2$

e) $E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

f) $E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

Step-by-step explanation:

We have the following distribution

x 0 1 2 3 4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

$P(X=3) = 0.1$

Part b

We want this probability:

$P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

Part c

For this case we want this probability:

$P(X=4) = 0.3$

Part d

$P(X=0) = 0.2$

Part e

We can find the mean with this formula:

$E(X)= \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

Part f

We can find the second moment with this formula

$E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

4 0

4 years ago

In circle A shown below, the measure of ∠BAD is 148°: Circle A with angle BAD measuring 148 degrees; points B, C, D lie on Circl

german

Answer:

52°

Step-by-step explanation:

In the figure attached, Circle A is shown. There we can see that:

∠BC + ∠CD = ∠BAD

We know that Arc BC measures 96° and arc BAD measures 148°. Replacing this data into the equation and solving for Arc CD, we get:

96° + ∠CD = 148°

∠CD = 148° - 96°

∠CD = 52°

5 0

3 years ago