Ask question

Ask question

All categories

3 years ago

14

Help i need help ASAP

1 answer:

Alexxx [7]3 years ago

5 0

Jane : 6

Lane : 3

Neesa : 213

So I think Jane is 6 inches tall or 1-1/5

My answer can be incorrect also so refer to Chrome too.

You might be interested in

Issac fills up his 20 -gallon gas tank with fuel that costs $3.54 per gallon. The expression 3.54g represents the cost for a num

Ray Of Light [21]

G represents the number of gallons

6 0

4 years ago

Read 2 more answers

Urgent I don't have time pllllllllzzzzzzzzzz<br> thhhxx

Elis [28]

I think its the first one

4 0

3 years ago

Read 2 more answers

The decimal number 11.8 can be read as "eleven and eight tens."<br><br><br> True or False

Semmy [17]

I mean it could be read that way but that's the way most people read fractions. 11 and 8/10. Most people I know read decimals like that as 11 point 8. So yes I suppose it's true; it's not wrong, but most people don't say it like that. And I don't know how they taught you to read it in your math course.

3 0

4 years ago

Read 2 more answers

What is the range of the function on the graph?

Ainat [17]

Check the picture below.

includes 2, notice, it doesn't go below 2 over the y-axis, and then it keeps on going up. So, the range is from +2 onwards.

4 0

4 years ago

Read 2 more answers

Historical data for a local manufacturing company show that the average number of defects per product produced is 2. In addition

charle [14.2K]

Answer:

The probability that there will be a total of 7 defects on four units is 0.14.

Step-by-step explanation:

A Poisson distribution describes the probability distribution of number of success in a specified time interval.

The probability distribution function for a Poisson distribution is:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...$

Let <em>X</em> = number of defects in a unit produced.

It is provided that there are, on average, 2 defects per unit produced.

Then in 4 units the number of defects is, $(2\times4)=8$ .

Compute the probability of exactly 7 defects in 4 units as follows:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14$

Thus, the probability of exactly 7 defects in 4 units is 0.14.

8 0

3 years ago