If 150g of sugar is used for 5 cakes
How much is used for 7 cakes?:)

Answers

Answer 1
Answer: Well, if 150 grams is used for 5 cakes, we can divide that.
150/5=30. 30 grams of sugar are used for each cake
You want to know how many grams for 7 cakes
30x7=210 grams of sugar are used for seven cakes.
Hope this helps!
Answer 2
Answer: 150 g - 5 cakes
x g - 7 cakes

x=(7\cdot150)/(5)\nx=210


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Joshua was offered a job that paid a salary of $77,500 in its first year. The salary was set to increase by 2% per year every year.If Joshua worked at the job for 5 years, what was the total amount of money earned over the 5 years, to the nearest wholenumber?
An angle that is greater than 90 degreesis called a _______.
Find the vertex of a function f(x)=-3x^2+9x-15 by completing the square
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In a certain math class, it is known that 18 students LOVE math review, and 5 students dislike math review. The teacher randomly selects three students from the class. What is the probability that all three students love math review?

Answers

The probability that the first random choice will be a matho-file and not a -phobe is 18/23. The probability of the second one is 17/22, and for the third one it's 16/21. The probability of all 3 is 4,896/10,626 = (18x17x16)/(23x22x21). That's 46.1% (rounded).

If Sean knows the slope of the line is 4 in the line passes through the point (1, negative 8 princesses, which equation should he use to find the Y intercept and what is the y-intercept of the line

Answers

Answer:

Step-by-step explanation:

y + 8 = 4(x - 1)

y + 8 = 4x - 4

y = 4x - 12

Question 2 (Multiple Choice Worth 6 points)(09.03 LC)

Tanya drives to work every day and passes two independently operated traffic lights. The probability that both lights are red is 0.55. The probability that the first light is red is 0.69. What is the probability that the second light is red, given that the first light is red?
0.83
0.75
0.80
0.73

Answers

Answer:

(C) 0.80

Step-by-step explanation:

Let the probability that the first light is red is= P_(1) and

the probability that the second light is red= P_(2). Then,

Probability that both lights are red= P( P_(1) and P_(2) )= 0.55,

Probability that the first light is red= P(P_(1))=0.69,

Probability that the second light is red, given that the first light is red=P( P_(2) /P_(1))=(P(P_(1) and P_(2)) )/(P(P_(1)) )

=(0.55)/(0.69)

=0.7971

0.80


Given:
Probability that both lights are red 0.55
Probability that 1st light is red 0.69

Find the probability that the 2nd light is red, given that the first light is red.

0.55/0.69 = 0.797 or 0.80

The probability that the 2nd light is red while the 1st light is also red is 0.80.

Mason has to mow 6 lawns today. So far, he has mowed 1 1/2 of them. How many does he have left to do?

Answers

He has left 4 1/2 lawns to mow. As there are total 6 lawns, Mason has mowed 1 1/2 lawns. The remaining lawns are 6 - 1 1/2 = 6 - 1.5 = 4.5 = 4 1/2

Answer:

4 1/2 is the answer! Hope this helps!

Step-by-step explanation:

Subtract. Write your answer in simplest form. 2/3 -3/8

Answers

2/3 = 16/24
3/8 = 9/24

16/24 – 9/24 = 7/24 

Your answer is 7/24

Answer:

Your answer is 7/24

Step-by-step explanation:

Which statement best describes why there is no real solution to the quadratic equation y = x^2 - 6x + 13?The value of (-6)2 - 4 • 1 • 13 is a perfect square.
The value of (-6)2 - 4 • 1 • 13 is equal to zero.
The value of (-6)2 - 4 • 1 • 13 is negative.
The value of (-6)2 - 4 • 1 • 13 is positive.

Answers

Answer:

The value of (-6)2 - 4 • 1 • 13 is negative.

Step-by-step explanation:

To solve a quadratic equation we can solve by factoring, graphing or through the quadratic formula. The formula is as follows:

x=\frac{-b+/-\sqrt{b^(2)-4ac } }{2a}

When the square root value is less than 0, a real solution cannot be found. This means when b^(2)-4ac<0. For the equation, a=1, b=-6 and c=13.

We substitute and simplify.

b^(2)-4ac=(-6)^2-4(1)(13)=36-4(13)=36-52=-16

This gives us an imaginary solution because the value of (-6)2 - 4 • 1 • 13 is negative.

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