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1 year ago

NEED ANSWER QUICKLY PLEASE

Mathematics

1 answer:

Sliva [168]1 year ago

6 0

(3x -11) + 44 + (4x-7) = 180

3x - 11 + 44 + 4x - 7 = 180 like terms

7x +26 = 180

7x = 154

X = 22

Angle 1
3(22) -11
55
Angle 2
4(22) -7
81
So the largest angle is 81 degrees!

Hope this helps!
Brainliest is much appreciated!

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b) $2y+\frac{1}{2}y=\frac{5y}{2}$

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3 years ago

The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with a mean of 12.1 fluid ounc

Ludmilka [50]

Answer:

$\mu = 12.1\\\sigma = 0.3$

a) Find the probability that the drink is less than 11.9 fluid ounces

We are supposed to find P(x<11.9)

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

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Hence the probability that the drink is less than 11.9 fluid ounces is 0.2514

b)Find the probability that the drink is between 11.6 and 11.9 fluid ounces

x= 11.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.6-12.1}{0.3}$

$z=-1.67$

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

We are supposed to find P(11.6<x<11.9)

So, P(11.6<x<11.9)= P(x<11.9)-P(x<11.6)

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

Refer the z table

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

P(-1.67<z<-0.67)= 0.2514-0.0475

P(-1.67<z<-0.67)= 0.2039

So, the probability that the drink is between 11.6 and 11.9 fluid ounces is 0.2039.

c)Find the probability that the drink is more than 12.6 fluid ounces

We are supposed to find P(x>12.6)

x= 12.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{12.6-12.1}{0.3}$

$z=1.67$

Refer the z table

P(x>12.6)=1-P(x<12.6)

P(z>1.67)=1-P(z<1.67)

P(z>1.67)=1-0.9525

P(z>1.67)=0.0475

Hence the probability that the drink is more than 12.6 fluid ounces is 0.0475

d) Is the drink containing more than 12.6 fluid ounces an unusual event?

Since p value is less than alpha (0.05)

Answer = No, because the probability that a drink containing more than 12.6 fluid ounces is less than 0.05, this event is not unusual.

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<h2>Answer with explanation:</h2>

Let p be the population proportion of parents who had children in grades K-12 were satisfied with the quality of education the students receive.

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Confidence interval : $\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

$0.44\pm (1.96)\sqrt{\dfrac{0.39(1-0.39)}{1055}}\\\\0.44\pm0.0299536805135\\\\0.44\pm0.03\\\\=(0.44-0.03, 0.44+0.03)\\\\=(0.41,\ 0.47)$

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