Scientific notation 0.000003408 =

James buys x books at a used-book sale. Enter an inequality that can be used to find the number of books James buys (x) when the

den301095 [7]

2x<15

Step-by-step explanation:

2×15=30 if he spends less than he would have to be below 15 please give brainliest

4 0

3 years ago

Between which two ordered pairs does the graph of f(x) = 1/2x2 + x – 9 cross the negative x-axis? (–6, 0) and (–5, 0) (–4, 0) an

lbvjy [14]

Answer:

Between (-6,0) and (-5,0).

Step-by-step explanation:

When it crosses the x axis, it's an x-intercept. To find the x-intercept, set y (aka f(x)) to zero. The solutions to the equation are approximately 3.36 and - 5.36. Since you're finding the negative x-intercept, you only need the negative solution - 5.36. This is in between (-6,0) and (-5,0).

3 0

3 years ago

The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0

3 years ago

A famous basketball player scored 97 points in a high scool game, breaking a national record. She made 43 more 2-point baskets t

ziro4ka [17]

X - 3pt baskets (he reaches 3x points)
y - 2pt baskets (he reaches 2y points)

We know, that y=x+43

97 - 1 free throw = 97 - 1 = 96
Sum of points: 96

3x + 2y = 96. Substitute to "y" an expression "x+43" :

3x + 2(x+43)=96
3x+2x+86=96
5x=10 |:5
x=2

So y=x+43=2+43=45

He has made 2 three points baskets and 45 two points baskets

8 0

3 years ago

Help first one gets brainllest

Vaselesa [24]

Answer:

What is the question???

3 0

3 years ago

Read 2 more answers