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3 years ago

I need help and will give braimleiest if right plus 25 points please

Mathematics

2 answers:

Bess [88]3 years ago

7 0

Hola aquí va la respuesta!!

$37°$

12345 [234]3 years ago

3 0

Answer:

I think <2 = 37⁰

Step-by-step explanation:

if I remember correctly, if its straight across from it, it must be the same angle.

which means that 1 and 3 must have the same angle, and 2 must have the same angle as the other side, meaning its angle is 37⁰.

You might be interested in

5x − 8( 1 + 7x ) = −8

xxTIMURxx [149]

Answer:

x = 0

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

Step-by-step explanation:

Step 1: Define equation

5x - 8(1 + 7x) = -8

Step 2: Solve for x

Distribute -8: 5x - 8 - 56x = -8
Combine like terms: -51x - 8 = -8
Isolate x term: -51x = 0
Isolate x: x = 0

Step 3: Check

Plug in x into the original equation to verify it is a solution.

Substitute in x: 5(0) - 8(1 + 7(0)) = -8
Multiply: 0 - 8(1 + 0) = -8
Add: -8(1) = -8
Multiply: -8 = -8

Here we see that -8 does indeed equal -8.

∴ x = 0 is a solution of the equation.

6 0

3 years ago

Selena's family went on a trip. The total hotel bill was $659 . The cost of the airfare was $633. use mental math to find the to

Vilka [71]

659+633

=1292

hope this helps

4 0

4 years ago

Read 2 more answers

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

What’s 3х + 7>4х +2

Kipish [7]

Answer:

$x < 5$