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3 years ago

What's the product of 70x6000

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

4 0

The product of 70 and 6000 is 420,000.

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mikes deli just sold 328 ham sandwiches this week. around how many ham sandwiches did mike sold every day this week?

Alenkasestr [34]

Answer:

46.9

Step-by-step explanation:

328/7 = 46.9

To double check: 46.9 x 7 = 328

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2 years ago

To make 5 blueberry pies,Mindy needs about 2 pounds of blueberries. How many pounds of blueberries does Mindy need to make 20 bl

Vsevolod [243]

20 lol sorry if it’s wrong

6 0

2 years ago

Read 2 more answers

Irene has 1 gallon of milk. She uses

Umnica [9.8K]

Answer:

Step-by-step explanation:

In 1 gallon, there are 128 ounces.

Divide 128 by 4 and get 32.

5 0

2 years ago

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic

Rina8888 [55]

Answer:

The probability that none of the LED light bulbs are defective is 0.7374.

Step-by-step explanation:

The complete question is:

What is the probability that none of the LED light bulbs are defective?

Solution:

Let the random variable X represent the number of defective LED light bulbs.

The probability of a LED light bulb being defective is, P (X) = p = 0.03.

A random sample of n = 10 LED light bulbs is selected.

The event of a specific LED light bulb being defective is independent of the other bulbs.

The random variable X thus follows a Binomial distribution with parameters n = 10 and p = 0.03.

The probability mass function of X is:

$P(X=x)={10\choose x}(0.03)^{x}(1-0.03)^{10-x};\ x=0,1,2,3...$

Compute the probability that none of the LED light bulbs are defective as follows:

$P(X=0)={10\choose 0}(0.03)^{0}(1-0.03)^{10-0}$

$=1\times 1\times 0.737424\\=0.737424\\\approx 0.7374$

Thus, the probability that none of the LED light bulbs are defective is 0.7374.

6 0

2 years ago

What is the radical form of the expression (2x^4y^5)^3/8

s344n2d4d5 [400]

We are given expression: $(2x^4y^5)^{3/8}$

Let us distribute 3/8 over exponents in parenthesis, we get

$(2^{3/8}x^{4\times 3/8}y^{5\times 3/8}) = (2^{3/8}x^{12/8}y^{15/8})$

$= (2^{3/8}x^{1\frac{4}{8}} y^{1\frac{7}{8}} )$

We can get x and y out of the radical because, we get whlole number 1 for x and y exponents for the mixed fractions.

So, we could write it as.

$(2^{3/8}x^{1\frac{4}{8}} y^{1\frac{7}{8}} ) = xy(2^{\frac{3}{8} }x^{\frac{4}{8}} y^{\frac{7}{8}} )$

Now, we could write inside expression of parenthesis in radical form.

$xy\sqrt[8]{2x^{3}x^4y^7}$

8 0

3 years ago