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3 years ago

7

Round 62,985 to the nearest ten

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!}
$

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!}
$

$\simeq 0.7408182207$ ------------------------------------(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!}
$

$\simeq 0.2222454662$ ------------------------------------(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