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3 years ago

Sally types 60 words per minute while

1 answer:

7 0

Sally types faster than sue per minute by 10 words. If you divide 3000 by 60 because there are 60 minutes in an hour you will get how many words she types a minute which would be 50.

You might be interested in

Use the theorem and the factor theorem to determine whether (y - 3) is a factor of (y^4+2y^2-4)

I am Lyosha [343]

Answer:

let p(x) = y^4 + 2y^2 - 4

and g(x) = y-3

=> p(x) / g(x) [ / means divide]

=> remainder = 8

therefore using both the theorem, the answer is C

7 0

3 years ago

Can anyone tell me how to work out 4/7 of 5 ????

HACTEHA [7]

Here, 4/7 * 5
Multiply numerators together, and denominators together
It would be: 4 * 5 / 7 * 1 = 20/7

So, your final answer is 20/7

Hope this helps!

7 0

4 years ago

Read 2 more answers

Using only the values given in the table for the function f(x) = x^3 - 3x - 2 what is the interval of x-values over which the fu

Nataly [62]

Answer:

3 = 3(x² - 1) We know, any Function f(x) decrease s in interval [a, b] when f'(x) < 0 ... -1 < x < 1. Hence, function f(x) = x³ - 3x - 2 , is decreased in (-1, 1).

4 0

3 years ago

A stack of cards consists of seven red and four blue cards. A second stack of cards consists of eleven red cards. A stack is sel

Ann [662]

Answer:

0.1733 = 17.33% probability the first stack was selected.

Step-by-step explanation:

To solve this question, it is needed to understand conditional probability, and the hypergeometric distribution.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Conditional probability:

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Probability of all red cards for the first stack:

For this, we use the hypergeometric distribution, as the cards all chosen without replacement.

7 + 4 = 11 cards, which means that $N = 11$ .

7 red, which means that $k = 7$

3 are chosen, which means that $n = 3$

We want all red, so we find P(X = 3).

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,11,3,7) = \frac{C_{7,3}*C_{4,0}}{C_{11,3}} = 0.2121$

Conditional probability:

Event A: All red

Event B: From the first stack.

Probability of all red cards:

0.2121 of 50%(first stack)

1 of 50%(second stack). So

$P(A) = 0.2121*0.5 + 1*0.5 = 0.60605$

Probability of all red cards and from the first stack:

0.21 of 0.5. So

$P(A \cap B) = 0.21*0.5 = 0.105$

What is the probability the first stack was selected?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.105}{0.60605} = 0.1733$

0.1733 = 17.33% probability the first stack was selected.

4 0

3 years ago

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 418 gram setting. It

VMariaS [17]

Answer: There is sufficient evidence to support the claim that the bags are under-filled.

Step-by-step explanation:

Since we have given that

$H_0:\mu=418\\\\H_a:\mu$

Sample mean = 411

Standard deviation = 20

n = 9

So, the test statistic value is given by

$z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\\z=\dfrac{411-418}{\dfrac{20}{\sqrt{9}}}\\\\\\z=\dfrac{-7}{\dfrac{20}{3}}\\\\\\z=-1.05$

At 0.025 level of significance,

critical value z = -2.306

since -2.306<-1.05

so, we will reject the null hypothesis.

Yes, there is sufficient evidence to support the claim that the bags are underfilled.

5 0

3 years ago