All categories

3 years ago

Solve for the missing number, founding to three decimal places when necessary b. 0.76 = 1425

Mathematics

1 answer:

boyakko [2]3 years ago

5 0

Answer:

1.425

Step-by-step explanation:

b.0.76

0.76÷1000=0.076

l425÷1000=1.425

You might be interested in

The african bush elephant weighs between 4.4 tons and 7.7 tons. What are its least and greatest wieghts rounded to the nearest t

WARRIOR [948]

4.4=4 tons rounded

7.7=8 tons rounded

6 0

4 years ago

Read 2 more answers

What is the value of y in the product of powers below?<br> 8^3 • 8^5 • 8^y = 8^-2 = 1/8^2

vodka [1.7K]

Answer:

y=0

Step-by-step explanation:

8 0

3 years ago

A local farm grows strawberries, blueberries, and raspberries.

e-lub [12.9K]

A)$30.98
B) 35.25
Step by step explanation:

3 0

3 years ago

Read 2 more answers

Greg needs to buy plastic spoons. Brand A has a box of 25 spoons for $1.99 . Brand B has a box of 42 spoons for $3.89 . Find the

Helga [31]

To find the price per spoon of brand A, divide the cost by the amount of spoons:

1.99 / 25 = 0.0796

Repeat this process for brand B:

3.89 / 42 = 0.0926

Brand A, at $0.08 a unit, is the better buy of brands A and B.

Hope this helps.

8 0

3 years ago

Read 2 more answers

Starting at 6 a.m. every morning, Matilda receives text messages on her cell phone from her mother, her best friend, and her bro

Dafna1 [17]

Answer:

a) 0.0013 = 0.13% probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

b) $1.6 \times 10^{-9}$ probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

Step-by-step explanation:

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A) Find the probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

Two from the best friend:

Her best friend sends a message once every 10 minutes.

From 6 to 7:30, there is an hour and a half, that is, 90 minutes, so the mean for her best friend is $\mu = \frac{90}{10} = 9$

Two messages is P(X = 2). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-9}*9^{2}}{(2)!} = 0.0050$

Two from the mother:

Message every hour = 60 minutes. So $\mu = \frac{90}{60} = 1.5$ . This is P(X = 2).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-1.5}*1.5^{2}}{(2)!} = 0.2510$

Two of her best friend and two of her mother:

Independent events, so the probability of both happening is the multiplication of their separate probabilities.

$p = 0.005*0.251 = 0.0013$

0.0013 = 0.13% probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

B) With a chance of 75% a text message contains a typo independent of the sender. Find the probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

In 3 hours, she is expected to receive:

3*60/10 = 18 messages from her best friend.

3*60/60 = 3 messages from her mother.

3*60/30 = 6 messages from her brother.

In total, 27 messages.

75% probability of a typo, so $\mu = 0.75*27 = 20.25$

This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-20.25}*20.25^{0}}{(0)!} = 1.6 \times 10^{-9}$

$1.6 \times 10^{-9}$ probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

8 0

3 years ago