All categories

3 years ago

Based on the line plot how many recipes call for more than 1/4tsp of salt

Mathematics

1 answer:

yulyashka [42]3 years ago

5 0

Do you think you could maybe insert a picture of the line plot?

Thanks :)

You might be interested in

Please help me Find X

solmaris [256]

The labeled angle is what's called an inscribed angle. There's a theorem (fittingly called the inscribed angle theorem) that says the measure of this angle is half the measure of the circular arc that it subtends.

In this case, the angle is subtended by an arc measuring 180°. So

(4x + 6)° = 1/2 • 180°

4x + 6 = 90

4x = 84

x = 21

7 0

1 year ago

A gallon of gas costs $2.93 per gallon. Marcie's car holds 10 gallons. If her tank is empty, how many will it cost to fill it?

melamori03 [73]

The answer is $29.3 because you multiply 2.93 times the ten gallons of gas.

6 0

2 years ago

Consider a local community of a rock chucks in Wyoming that are increasing at a rate of 11% per month. Using the Rule of 70, ans

natka813 [3]

I am not good at math or I would help :/

8 0

3 years ago

Determine the length of the unknown sides for the right triangle

alisha [4.7K]

Answer:

Below in bold.

Step-by-step explanation:

sin 66.4 = opposite side / hypotenuse

sin 66.4 = 21.3 / y

y = 21.3 / sin 66.4

y = 23.24.

tan 66.4 = 21.3 / x

x = 21.3 / tan 66.4

x = 9.31.

3 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago