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3 years ago

8993.2x 10³ use the powers of ten rules please help iv been stuck on this

Mathematics

1 answer:

vladimir2022 [97]3 years ago

6 0

Answer:

I think the answer is 8993200

Step-by-step explanation:

i think its wrong but I try my best

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How do you find the quadratic function with vertex (-2,5) and point (0,9)?

Anastaziya [24]

Answer: y = (x +2)² + 5

Step-by-step explanation:

y = a(x - h)² + k where "a" is the leading coefficient and (h, k) is the vertex

Since we don't know "a", we need to plug in the point (x, y) and the vertex (h, k) to solve for "a": (x, y) = (0, 9) and (h, k) = (-2, 5)

y = a(x - h)² + k

9 = a(0 - (-2))² + 5

9 = a(0 + 2)² + 5

9 = a(2)² + 5

-5 -5

4 = a(4)

÷4 ÷4

1 = a

Next, plug in "a" and the vertex (h, k):

y = a(x - h)² + k

y = 1(x +2)² + 5

y = (x +2)² + 5

6 0

4 years ago

1/2*(6*1+7)+11 show the result

fomenos

Well I got 17.5
You would start with 6x1=6
Then you would add 7: 6+7=13
Next you would multiply 13 by (1/2): 13(1/2)=6.5
Finally you would add 11 to 6.5: 6.5+11= 17.5

3 0

3 years ago

12 - 3/4 (d + 16) = negative 5

Nutka1998 [239]

Answer: d=20/3

Step By Step:

Step 1, Simplify:

12-3/4(d+16)=-5

(-3/4)d+(12-12)=-5

(-3/4)d=-5

Step 2, Multiply each side by 4/(-3)

(4/-3)*(-3/4)d=(4/-3)*-5

d=20/3

8 0

3 years ago

Read 2 more answers

Quiz

ivanzaharov [21]

Answer: Its A, Your Welcome.

Step-by-step explanation:

4 0

4 years ago

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introvert

Setler [38]

Answer:

$P(x \ge 5) = 1.000$ ---- At least 5 from marketing departments are extroverts

$P(x=15) = 0.013$ ---- All from marketing departments are extroverts

$P(x = 0) = 0.002$ ---------- None from computer programmers are introverts

Step-by-step explanation:

See comment for complete question

The question is an illustration of binomial probability where

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$(a):\ P(x \ge 5)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

Using the complement rule, we have:

$P(x \ge 5) = 1 - P(x < 5)$

So, we have:

$P(x < 5) =P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)$

$P(x = 0) = ^{15}C_{0} * (75\%)^0 * (1 - 75\%)^{15 - 0} = 1 * 1 * (0.25)^{15} = 9.31 * 10^{-10}$

$P(x = 1) = ^{15}C_{1} * (75\%)^1 * (1 - 75\%)^{15 - 1} = 15* (0.75)^1 * (0.25)^{14} = 4.19 * 10^{-8}$

$P(x = 2) = ^{15}C_{2} * (75\%)^2 * (1 - 75\%)^{15 - 2} = 105* (0.75)^2 * (0.25)^{13} = 8.80 * 10^{-7}$

$P(x = 3) = ^{15}C_{3} * (75\%)^3 * (1 - 75\%)^{15 - 3} = 455* (0.75)^2 * (0.25)^{12} = 0.0000153$

$P(x = 4) = ^{15}C_{4} * (75\%)^4 * (1 - 75\%)^{15 - 4} = 1365 * (0.75)^4 * (0.25)^{11} = 0.000103$

So, we have:

$P(x < 5) = (9.31 * 10^{-10}) + (4.19 * 10^{-8}) + (8.80 * 10^{-7}) + 0.0000153 + 0.000103$

$P(x < 5) = 0.00011922283$

Recall that:

$P(x \ge 5) = 1 - P(x < 5)$

$P(x \ge 5) = 1 - 0.00011922283$

$P(x \ge 5) = 0.9998$

$P(x \ge 5) = 1.000$ --- approximated

$(b)\ P(x = 15)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x=15) = ^{15}C_{15} * 0.75^{15} * (1 - 0.75)^{15-15}$

$P(x=15) = 1 * 0.75^{15} * (0.25)^{0$

$P(x=15) = 0.013$

$(c)\ P(x = 0)$

$n=5$ ---------- computer programmers

$p = 70\%$ --- proportion that are introverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}$

$P(x = 0) = 1 * 1 * (0.30)^5$

$P(x = 0) = 0.002$

3 0

3 years ago

8993.2x 10³ use the powers of ten rules please help iv been stuck on this​

8993.2x 10³ use the powers of ten rules please help iv been stuck on this