What is (4+49)-4x10=

s2008m [1.1K]

-42.2
i hope this helps!

8 0

2 years ago

I was at the store, and saw two sizes of avocados being sold. The smaller size sold for $0.92 each. For what prices would the la

Firdavs [7]

Answer:

Large avocados should cost $ 1.83 or less to be a good deal.

Step-by-step explanation:

Since there are two types of avocado in the store, some small at $ 0.92 and others larger, to determine at what price large avocados would be a good deal, an equivalence must be established in this regard:

Thus, if two small avocados are equal to one large, buying two small avocados at $ 0.92 the total price would be $ 1.84. Therefore, any large avocado that sells for less than $ 1.84 would be a good deal. Thus, large avocados should cost $ 1.83 or less to be a good deal.

3 0

2 years ago

Write an equation of a polynomial with the given characteristics: a quadratic function has x -intercepts of -3 and 1, and a y-in

fredd [130]

Answer:

f(x) = -3(x+3)(x-1)

Step-by-step explanation:

x = -3 & 1; f(x) = 9

f(x) = a(x-r1)(x-r2)

f(0) = a(x-r1)(x-r2) = 9; 9 = a(0-(-3))(0-1)

9 = a(3)(-1); 9 = a(-3)

a = -3

f(x) = -3(x+3)(x-1)

5 0

2 years ago

Read 2 more answers

he number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.02

Yakvenalex [24]

Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

$P(k)= \frac{0.2^ke^{-0.2}}{k!}$

a)

In this case, we are looking for P(0)

$P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%$

So, the probability that a car has no flaws is 98.01%

b)

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and

$P(k)= \frac{2^ke^{-2}}{k!}$

and

$P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%$

And the probability that 10 cars have no flaws is 13.53%

c)

Here, we are looking for P(1) with P defined as in b)

$P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%$

Hence, the probability that at most one car has no flaws is 27.06%

6 0

3 years ago

Read 2 more answers

A book store has twice as many history books as science books. The store has 81 history and science books altogether how many of

atroni [7]

let h= history books

science books = s

h=2s there are 2 times as many history as science so to get them equal we double the science

h+s=81 there are 81 total history and science books

2s+s=81 substitute 2s for h in the above equation

3s = 81 combine like terms

s = 27

There are 27 science books

h=2s

h=2(27)

h=54

there are 54 history books

4 0

3 years ago