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lana [24]
2 years ago
9

Kendra is also buying souvenirs for the family reunion. She wants to spend under $2.25 for each item.

Mathematics
1 answer:
PIT_PIT [208]2 years ago
5 0

Answer:

1st,2nd,4th,

Step-by-step explanation:

You have to divide the price by the numbers of items!

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1 cm equals 10 mm how many millimeters does 2.5 cm equal label your answer ​
kodGreya [7K]
Answer: 25 mm

Explanation:

Multiply 2.5 by 10 since 1=10 and there are 2.5

2.5(10)
=25
3 0
2 years ago
The number of assists per match for the setter on your school's volleyball team has a mean of 58 and a standard deviation of 7.
serious [3.7K]
To get the number of standard deviation that 77 is from the mean, we get the z-score:
z-score is given by
z=(x-μ)/σ
where:
μ-mean
σ-standard deviation
thus the value of z from the information above is:
x=77
μ=58
σ=7
z=(77-58)/7=2.7143

3 0
3 years ago
An unfair coin is twice as likely to land "Heads" as "Tails". If the coin is tossed twice, what is the probability that it will
Ierofanga [76]
The probability of getting two heads on two coin tosses is 0.5 x 0.5 or 0.25. A visual representation of the toss of two coins. The Product Rule is evident from the visual representation of all possible outcomes of tossing two coins shown above.

Answer / 1/2
7 0
1 year ago
Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o
Brrunno [24]

Answer:

AB = \sqrt{a^2 + b^2-2abCos\ C}

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

Sin\ C = \frac{h}{a}

Make h the subject:

h = aSin\ C

Also, in BOC (Using Pythagoras)

a^2 = h^2 + x^2

Make x^2 the subject

x^2 = a^2 - h^2

Substitute aSin\ C for h

x^2 = a^2 - h^2 becomes

x^2 = a^2 - (aSin\ C)^2

x^2 = a^2 - a^2Sin^2\ C

Factorize

x^2 = a^2 (1 - Sin^2\ C)

In trigonometry:

Cos^2C = 1-Sin^2C

So, we have that:

x^2 = a^2 Cos^2\ C

Take square roots of both sides

x= aCos\ C

In triangle BOA, applying Pythagoras theorem, we have that:

AB^2 = h^2 + (b-x)^2

Open bracket

AB^2 = h^2 + b^2-2bx+x^2

Substitute x= aCos\ C and h = aSin\ C in AB^2 = h^2 + b^2-2bx+x^2

AB^2 = h^2 + b^2-2bx+x^2

AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2

Open Bracket

AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C

Reorder

AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C

Factorize:

AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C

In trigonometry:

Sin^2C + Cos^2 = 1

So, we have that:

AB^2 = a^2 * 1 + b^2-2abCos\ C

AB^2 = a^2 + b^2-2abCos\ C

Take square roots of both sides

AB = \sqrt{a^2 + b^2-2abCos\ C}

6 0
3 years ago
S+4t= r for s<br> Please I need help!!!!!
kenny6666 [7]
S + 4t = r
subtract 4t from both sides
s = r - 4t  answer
6 0
3 years ago
Read 2 more answers
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