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2 years ago

Solve the following equation: 7x– 9=19

Mathematics

2 answers:

Gelneren [198K]2 years ago

4 0

Answer:

x=4

Step-by-step explanation:

7x - 9 = 19

add 9 to both sides

7x = 28

divide by 7

x = 4

Sladkaya [172]2 years ago

4 0

Answer:

$x=4$

Step-by-step explanation:

First, you take the equation and add 9 to both sides ( $7x-9+9=19+9$ ) which will cancel out the 9 on the left side, and add onto the 19 which would be 28.

Next, you simplify the answer of that to get $7x=28$

After you divide 7 from both sides ( $\frac{7x}{7}=\frac{28}{7}$ )

Finally, after solving that division problem you will get the answer of the equation which is $x=4$

Brainliest pls?

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Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 w

irina1246 [14]

Answer:

A 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students who eat cauliflower

n = sample of students

p = population proportion of students who eat cauliflower

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

Now, in Agresti and Coull's method; the sample size and the sample proportion is calculated as;

$n = n + Z^{2}__(\frac{_\alpha}{2})$

n = $24 + 1.96^{2}$ = 27.842

$\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_) }{2} }{n}$ = $\hat p = \frac{2+\frac{1.96^{2} }{2} }{27.842}$ = 0.141

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ , $0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ ]

= [0.012, 0.270]

Therefore, a 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95% confident that the proportion of students who eat cauliflower on Jane's campus is between 0.012 and 0.270.

7 0

3 years ago

PQ⊥PS , m∠QPR=7x−9, m∠RPS=4x+22 Find : m∠QPR

emmainna [20.7K]

Given:

It is given that,

PQ ⊥ PS and

∠QPR = 7x-9

∠RPS = 4x+22

To find the value of ∠QPR.

Formula

As per the given problem PR lies between PQ and PS,

So,

∠QPR+∠RPS = 90°

Now,

Putting the values of ∠RPS and ∠QPR we get,

$7x-9+4x+22 = 90$

or, $11x = 90-22+9$

or, $11x = 77$

or, $x = \frac{77}{11}$

or, $x = 7$

Substituting the value of $x = 7$ in ∠QPR we get,

∠QPR = $7(7)-9$

or, ∠QPR = $40^\circ$

Hence,

The value of ∠QPR is 40°.

3 0

3 years ago

Someone please help me with these 2 questions I will mark brainlist!!!!

Tamiku [17]

ANSWER for the first one:
A=a+b/2*h
A= 4.5+9/2*6
A= 40.5 cm2

ANSWER for the second one:
9*3= 27
3*3= 9
9*3= 27
A= 63 cm2

8 0

3 years ago

2. In a given population of two-earner male-female couples, male earnings have a mean of $40,000 per year and a standard deviati

Kobotan [32]

Answer: $85,000

Step-by-step explanation:

Given : In a given population of two-earner male-female couples, male earnings have a mean of $40,000 per year and a standard deviation of $12,000.

$\mu_M=40,000\ \ ;\sigma_M=12,000$