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OLEGan [10]
3 years ago
9

Whats 107 over 1000 as a decimal​

Mathematics
2 answers:
ivann1987 [24]3 years ago
6 0

Step-by-step explanation:

107 / 1000 = 0.107

Vladimir79 [104]3 years ago
6 0

Answer:0.107

Step-by-step explanation: 107 over 1000 is 107 decided by 1000, which is 0.107

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What is the GCF of 5 10 and 30
Greeley [361]

Answer:

5

Step-by-step explanation:

5 because it can only go into itself one time, 5 x 2 is 10 and 5 x 6 is 30

7 0
3 years ago
Consider two markets: the market for cat food and the market for dog food. The initial equilibrium for both markets is the same,
umka21 [38]

Answer:

elasticity supply of dog food = 2.61

elasticity supply of cat food = 1.71

Step-by-step explanation:

The midpoint formula for elasticity is:

Elasticity = \frac{(Q2-Q1)/[(Q2+Q1)/2]}{(P2-P1)/[(P2+P1)/2]}

Point 1: Q = 39.0 and P = 5.50

Point 2: Q = 101.0 and P = 7.75

Elasticity\ supply\ of\ dog\ food = \frac{(101.0-39.0)/[101.0+39.0)/2]}{(7.75-5.50)/[(7.75+5.50)/2]}=2.61

Doing the same for the cat food:

Elasticity\ supply\ of\ cat\ food = \frac{(71.0-39.0)/[71.0+39.0)/2]}{(7.75-5.50)/[(7.75+5.50)/2]}=1.71

4 0
4 years ago
Find the commutative property of multiplication
exis [7]

Answer:

they are all true they are just expressed different

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is logs(4.7)+logs 2 written as a single log?
Ipatiy [6.2K]

ANSWER

log_{s}(56)

EXPLANATION

The given logarithmic expression is:

log_{s}(4 \times 7)  +  log_{s}(2)

Recall and use the product rule of logarithm

log_{a}(b)  +  log_{a}(c)  =  log_{s}(bc)

We apply this rule to obtain,

log_{s}(4 \times 7)  +  log_{s}(2)  = log_{s}(4 \times 7 \times 2)

We multiply out the argument to get;

log_{s}(4 \times 7)  +  log_{s}(2)  = log_{s}(56)

The correct answer is

log_{s}(56)

7 0
4 years ago
At the beginning of year 1, Jonah invests $300 at an annual
olga55 [171]

At the beginning of year 6, that means 5 years only have elapsed, thus t = 5.

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$300\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases} \\\\\\ A=300\left(1+\frac{0.04}{1}\right)^{1\cdot 5}\implies A=300(1.04)^5

6 0
4 years ago
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