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3 years ago

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How many different 1-topping pizzas can be formed by choosing a size and topping if your choices for size are small, medium, lar

ge, and extra-large while your choices for toppings are pepperoni, sausage, onions, ham, green peppers, mushrooms and bacon?
a. 28
b. 18
c. 24
d. 11

2 answers:

Rudik [331]3 years ago

4 0

The answer is A. 28

bekas [8.4K]3 years ago

3 0

28.........................................

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The diameters of bolts produced in a machine shop are normally distributed with a mean of 6.4 millimeters and a standard deviati

dsp73

Answer:

The bolts with diameter less than 5.57 millimeters and with diameter greater than 5.85 millimeters should be rejected.

Step-by-step explanation:

We have been given that the diameters of bolts produced in a machine shop are normally distributed with a mean of 5.71 millimeters and a standard deviation of 0.08 millimeters.

Let us find the sample score that corresponds to z-score of bottom 4%.

From normal distribution table we got z-score corresponding to bottom 4% is -1.75 and z-score corresponding to top 4% or data above 96% is 1.75.

Upon substituting these values in z-score formula we will get our sample scores (x) as:

Therefore, the bolts with diameters less than 5.57 millimeters should be rejected.

Now let us find sample score corresponding to z-score of 1.75 as upper limit.

Therefore, the bolts with diameters greater than 5.85 millimeters should be rejected

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3 years ago

The position, x, of an object is given by the equation x = A + Bt +Ct2, where t refers to time. What are the dimensions of A, B,

Serjik [45]

Answer: Dimensions of A are of length [L]

Dimensions of B are of $LT^{-1}$

Dimensions of C are of $LT^{-2}$

Step-by-step explanation:

The given equation is

$x(t)=A+Bt+Ct^{2}$

Since the dimension on the L.H.S of the equation is [L] , each of the terms on the right hand side should also have dimension of length[L] to be dimensionally valid

Thus

Dimensions of A = [L]

Dimensions of Bt = [L]

$Bt=[L]\\\\$ $[B][T]=[L]$ $\\\\\therefore [B]=LT^{-1}$

Similarly

Dimensions of $Ct^{}2 = [L]$

$Ct^{2}=[L]\\\\[C][T]^{2}=[L]\\\\\therefore [C]=LT^{-2}$

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3 years ago

So soft ball bags are on clearance for an additional 30% off the sales price Finley finds one size:it was originally $70.00,but

KonstantinChe [14]

Answer:

$35

Step-by-step explanation:

50 * 30% = 15$ discount

50 - 15 = 35$ cost

30% discount is additional so based off sales price

5 0

3 years ago

(problem 6.13 page 97) Consider a population in which 80% of males and 60% of females are employed. In this population, 55% of i

svp [43]

Answer:

There is a 50.77% probability that no more than one of those chosen is not employed.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they will be employed, or they will not. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

We have these following percentages:

55% of the individuals are female.

So 45% of the individuals are male.

80% of males are employed. So 20% of males are unemployed.

60% of females are employed. So 40% of females are unemployed.

If I pick five persons at random from this population, what is the probability that no more than one of those chosen is not employed?

Using the binomial distribution, p is the probability that a person is unemployed. 40% of the females and 20% of the males are unemployed. The population is 55% females and 45% males. So

$p = 0.4*0.55 + 0.2*0.45 = 0.31$

There are five persons, so $n = 5$

What is the probability that no more than one of those chosen is not employed?

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.31)^{0}.(0.69)^{5} = 0.1564$

$P(X = 1) = C_{5,1}.(0.31)^{1}.(0.69)^{4} = 0.3513$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1564 + 0.3513 = 0.5077$

There is a 50.77% probability that no more than one of those chosen is not employed.

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3 years ago

Please help someone

uysha [10]

Answer:

96

Step-by-step explanation:

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3 years ago