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Tanya [424]
3 years ago
11

What is the answer to 64+(-15)+(-5)

Mathematics
2 answers:
MissTica3 years ago
8 0

Answer:

44

Step-by-step explanation:

MrRissso [65]3 years ago
8 0
The answer is 44

step by step explanation:
64+(-15)+(-5)

-15+(-5)= -20
64+(-20)= 44
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Find 20 percent of $20 by multiplying 0.20 x20

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HELP ME PLEASE WITH 1,2 AND 3 PLEASE DON'T TAKE LONG TO RESPOND !!​
soldi70 [24.7K]

Answer:

1. -19x

3. 23 -  8x

5. -18x + 2y

Step-by-step explanation:

first: 3x - x = 2x

2x - 22x = -20x

-20x + x = -19x

second one:

8 - (-15) = 23

12x - 20x = -8 x

23 - 8x

third:

-10x - 8x = -18x

3y - y = 2y

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6 0
3 years ago
Read 2 more answers
Write an expression that evaluates to true if the int associated with number_of_prizes is divisible (with no remainder) by the i
hodyreva [135]

We need to find the expression for " number_of_prizes is divisible number_of_participants". Also there should not remain any remainder left. On in order words, we can say the reaminder we get after division is 0.

Let us assume number of Prizes are = p and

Number of participants = n.

If we divide number of Prizes by number of participants and there will be not remainder then there would be some quotient remaining and that quotent would be a whole number.

Let us assume that quotent is taken by q.

So, we can setup an expression now.

Let us rephrase the statement .

" Number of Prizes ÷ Number of participants  = quotient".

p ÷ n = q.

In fraction form we can write

p/n =q   ; n ≠ 0.


4 0
3 years ago
What is 23 inches in centimetres?
Anettt [7]

Answer: 58.42

Formula: Multiply the length value by 2.54

23×2.54 =58.42

12 Inches = 30.48 Centimetres

7 0
3 years ago
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The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

4 0
3 years ago
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