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2 years ago

5

Please help I’m stuck

1 answer:

Nutka1998 [239]2 years ago

3 0

Answer:

96

Step-by-step explanation:

Its a pattern :) BTW can you help with mine please

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Orders for a computer are summarized by the optional features that are requested. The proportion of orders with no optional feat

Scorpion4ik [409]

Answer:

(a) 71.00%

(b) 63.00%

Step-by-step explanation:

No optional features = 0.29

One feature = 0.34

Two features = 0.37

(a) The probability that an order requests at least one optional feature is 100% minus the probability of an order requesting no optional features:

$P=1-0.29=0.71=71.00\%$

(b) The probability that an order does not request more than one optional feature is 100% minus the probability of an order requesting two optional features:

$P = 1 -0.37=0.63 = 63.00\%$

3 0

3 years ago

X≥1 x≤7 y≥-1/3x+6

Goshia [24]

Answer:

Step-by-step explanation:

i am unsure of what you actually need, but i want to help you.

if you were to set the boundaries of the feasibility region it would be as follows:

as points

A(1, 17/3) B(7, 11/3) C(7, infinity) D(1, infinity)

as graphs

y=-1/3x+6 where 1≤x≤7

x=1

x=7

7 0

3 years ago

Read 2 more answers

2(f-9)=-28 f=?<br>-3(2t+3)=-21 t=?<br>

SVETLANKA909090 [29]

Answer:

1. f= 3/5

2. t= 3/5

Step-by-step explanation:

FYI you can use the app photo math, you just take a pic of the problem and it gives you the answer and explains the steps and it is free.

5 0

2 years ago

Read 2 more answers

Find the inverseof the function:<br> y=(x – 3)^2

ollegr [7]

Answer:

f(x)= $\sqrt{x}$ +3, - $\sqrt{x}$ +3

Step-by-step explanation:

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Clarge%7B%5Cbold%20%5Cred%7B%20%5Csum%20%5Climits_%7B8%7D%5E%7B4%7D%20%7Bx%7D%5E%7B2%7D%2

kiruha [24]

Answer:

No solution

Step-by-step explanation:

We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

$\sum_{x=8}^{4}x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$

Note that for

$\sum_{x=a}^b f(x)$

it is assumed $a\leq x \leq b$ and in your case $\nexists x\in\mathbb{Z}: a\leq x\leq b$ for $a>b$

In fact, considering a set $S$ we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing$ that satisfy $S = S \cup \varnothing$

This means that, by definition $\sum_{x=a}^b \varnothing = 0$

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

For

$9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

we have other problems. Actually, this case is really bad.

Note that $\cos^2(\infty)$ has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$ , the cosine function oscillates between $[-1, 1]$ , and therefore it is undefined. Thus, we cannot evaluate

$9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

has no solution

7 0

2 years ago