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Lera25 [3.4K]
3 years ago
7

Round 62,985 to the nearest ten

Mathematics
2 answers:
kap26 [50]3 years ago
8 0
62,990
hope this helps
Zigmanuir [339]3 years ago
6 0

Answer:

63.000

Step-by-step explanation:

tell me if you need something pls branliest

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Describe how the product of 0.4 x 2 would look on a ten-square grid.
Gnesinka [82]
<h3>Answer:  8</h3>

Explanation:

Draw out a 10 squares.

Shade in 4 of them to represent 4/10 = 0.40 = 0.4

Then use another color to shade another 4 squares. You should have 4+4 = 8 squares shaded, out of 10 total, giving 8/10 = 0.80 = 0.8

This is one visual way to see how 0.4 x 2 = 0.8

In other words, 0.4 x 2 means we have two copies of 0.4 added together. You could write it as 0.4 x 2 = 0.4 + 0.4 = 0.8 if you wanted.

See the diagram below.

7 0
2 years ago
Find the product -a2 b2 c2 (a+b-c)
mina [271]
Use the distributive property:
a(c+b)=ab+ac
And
a^n\cdot a^m=a^{n+m}\\\\a=a^1

-a^2b^2c^2(a+b-c)=-a^2ab^2c^2-a^2b^2bc^2+a^2b^2c^2c=-a^3b^2c^2-a^2b^3c^2+a^2b^2c^3

7 0
3 years ago
What is the coefficient of q in the sum of (2/3q-3/4) and(-1/6q-r)?
Harlamova29_29 [7]

Answer:

The coefficient is \frac{1}{2}

Step-by-step explanation:

The given sum is

(\frac{2}{3}q -\frac{3}{4})+(-\frac{1}{6}q-r)

We can choose to add only the coeffients of q or simplify the whole sum.

=\frac{2}{3}q -\frac{1}{6}q-\frac{3}{4})-r)

We collect the LCM for the q terms;

=\frac{4-1}{6}q-\frac{3}{4})-r)

This will give us;

=\frac{3}{6}q-\frac{3}{4})-r)

Which is the same as

=\frac{1}{2}q-r-\frac{3}{4}

The coefficient is \frac{1}{2}

6 0
3 years ago
Read 2 more answers
What is the equation of a line that passes through the point (8, −2) and is parallel to the line whose equation is 3x + 4y = 15?
Dafna1 [17]

Step 1

<u>Find the slope of the given line</u>

we have

3x+4y=15

Isolate the variable y

Subtract 3x both sides

3x+4y-3x=15-3x

4y=-3x+15

Divide by 4 both sides

y=-\frac{3}{4}x+ \frac{15}{4}

the slope of the given line is

m=-\frac{3}{4}

Step 2

Find the equation of the line that passes through  the point (8,-2) and is parallel to the given line

we know that

if two lines are parallel, then their slopes are equal

The equation of the line into point-slope form is equal to

y-y1=m(x-x1)

we have

m=-\frac{3}{4}

(x1,y1)=(8,-2)

substitute in the equation

y+2=-\frac{3}{4}(x-8)

y=-\frac{3}{4}x+6-2

y=-\frac{3}{4}x+4  or  3x+4y=16

therefore

<u>the answer is</u>

y=-\frac{3}{4}x+4

or

3x+4y=16

8 0
3 years ago
Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1
alexandr1967 [171]

Answer:

a) \simeq 0.3012   b) \simeq 0.0494 c) \simeq 0.2438

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, \frac{1.2}{4}

= 0.3 collisions per  month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

          P(X =x) = \frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}&#10;

                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}&#10;

                      \simeq 0.7408182207 ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

     \simeq 0.3012  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}&#10;

                      \simeq 0.2222454662 ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

1 collision in 6 months period means

                                \frac{1}{6} collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6]  (which is to be estimated)

=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

\simeq 0.6543894283-------------------------------------------(6)

So,

P(1  collision in 6 month period)

  =  0.6543894283^{6}

   \simeq 0.0785267444 ------------------------------------------------(7)

So,

P(No collision in 6 months period)

  = (P(X =0)^{6}

   \simeq 0.1652988882 ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

\simeq  0.2438 ---------------------------------------------(9)          

7 0
3 years ago
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