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Semmy [17]
3 years ago
10

Write the value of the underlined digit. 6,581,678 _

Mathematics
1 answer:
aliya0001 [1]3 years ago
7 0
Its in the millions place. 6 million.
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Can someone help me
Lemur [1.5K]
Multiplying fractions is just multiplying the numerators together and the denominators together.

\sf\dfrac{4}{7}\times\dfrac{1}{4}\rightarrow\dfrac{4\times 1}{7\times 4}\rightarrow\dfrac{4}{28}\rightarrow\dfrac{1}{7}

\sf\dfrac{4}{7}\times\dfrac{1}{6}\rightarrow\dfrac{4\times 1}{7\times 6}\rightarrow\dfrac{4}{42}\rightarrow\dfrac{2}{21}

To find out which one is bigger, we need a common denominator. The LCM of 7 and 21 is 21, so convert the first fraction into a denominator of 21. 7 goes into 21 three times, multiply this to the numerator and denominator:

\sf\dfrac{1}{7}\rightarrow\dfrac{1\times 3}{7 \times 3}\rightarrow\dfrac{3}{21}

Now just compare the numerators to see which one is bigger.

\sf\dfrac{3}{21}>\dfrac{2}{21} therefore, \sf\dfrac{4}{7}\times\dfrac{1}{4}>\dfrac{4}{7}\times\dfrac{1}{6}
5 0
4 years ago
4 more than the quotient of 7 and m
tangare [24]

Answer:

4+ 7/m

Step-by-step explanation:

there might be parentheses but this is what the expression  would look like

3 0
3 years ago
Read 2 more answers
PLS HELP ASAP!!!<br> domain &amp; range of g(x)?
Morgarella [4.7K]
Domain: (-infinity, +infinity)
range: (0, +infinity)
6 0
3 years ago
If SAT scores are normally distributed with a mean of 1518 and a standard deviation of 325, find the probability that a randomly
aliya0001 [1]
Mean, x_bar = 1518
Standard deviation, sigma = 325
Range required: 1550 ≤ X ≤ 1575

Z = (X - x_bar)/sigma

Z1 = (1550-1518)/325 ≈ 0.1
Z2 = (1575-1518)/325 ≈ 0.18

From Z tables,
P(Z1) = 0.5398
P(Z2) = 0.5714

P(1550≤X≤1575) = P(Z2) - P(Z1) = 0.5714 - 0.5398 = 0.0316

The correct answer is C.
3 0
3 years ago
Exercise 10.10.2: Distributing a coin collection. About A man is distributing his coin collection with 35 coins to his five gran
11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

Given that:

all coins are same;

The same implies that the number of the non-negative integral solution of the equation:

x_1+x_2+x_3+x_4+x_5 = 35

x_1 > 0 ;  \ \ \ x_1 \  \varepsilon  \ Z

Thus, the number of the non-negative integral solution is:

^{(35+3-1)}C_{5-1} = ^{39}C_4

(b)

Here all coins are distinct.

So; the number of distribution appears to be an equal number of ways in arranging 35 different objects as well as 5 - 1 - 4 identical objects

i.e.

= \dfrac{(35+4)!}{4!}

= \dfrac{39!}{4!}

(c)

Here; provided that the coins are the same and each grandchild gets the same.

Then;

x_1+x_2+x_3+x_4+x_5 = 35

x_1 > 0 ;  \ \ \ x_1 \  \varepsilon  \ Z

x_1=x_2=x_3=x_4=x_5

5x_1 = 35\\\\ x_1= \dfrac{35}{5} \\ \\  x_1= 7

Thus, each child will get 7 coins

(d)

Here; we need to divide the 35 coins into 5 groups, this process will be followed by distributing the coin.

The number of ways to group them into 5 groups = \dfrac{35!}{(7!)^55!}

Now, distributing them, we have:

\mathbf{\dfrac{35!}{(7!)^55!}  \times 5!= \dfrac{35!}{(7!)^5}}

3 0
3 years ago
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