All categories

3 years ago

Place value of 3 in 5678.321

Mathematics

2 answers:

gregori [183]3 years ago

5 0

The place value of 3, is Tenths.

erma4kov [3.2K]3 years ago

4 0

The place value for the 3 is tenths

You might be interested in

NEED HELP ASAP!!!! Okay, I have 5 proportional relationship questions, PLEASE HELP:

Natasha_Volkova [10]

Answer:

The temperature yesterday was -6 because 7-13 is -6.

Step-by-step explanation:

8 0

3 years ago

Is 7s+8+5s equal to 12s+8

MatroZZZ [7]

Yes you can combine the 7s and the 5s to get 12s then add the plus 8

7 0

3 years ago

Read 2 more answers

When two 6-sided dice are rolled, there are 36 possible outcomes. Find the probability that the sum is 8

katrin2010 [14]

Answer:

5/36

Step-by-step explanation:

Probability is the outcome that a event will occur

Probability = Expected outcome/Total outcome

Since two sided dice are rolled, the total outcome will be expressed as;

Total outcome = 36

Since we are to find the probability that the sum is 8, the values that gives a sum of 8a re;

(2,6)(6,2), (5, 3), (3,5), (4, 4)

The expected outcome = 5

Probability = 5/36

Hence the probability that the sum is 8 is 5/36

4 0

3 years ago

The double box plot shows the test scores for Miss Robinson's second and fourth period

Gennadij [26K]

We can’t see the box plot?

4 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