All categories

3 years ago

Find the value of x?

Mathematics

1 answer:

Marrrta [24]3 years ago

5 0

Step-by-step explanation:

180-89=91(angles on a straight lline)

5x-24=91(Corresponding angles)

5x =91+24

5x =115

x =23 degrees

94+13x-5=180(allied angles)

94+13x-5 =180-94+5

13x =91

x=7

7(x-7)=7x-49

7x-49=126(corresponding angles)

7x =126+49

7x=175

x=25

Hope you got it

If you have any question just ask me

If you think this is the best answer please mark me as BRAINLIEST

You might be interested in

How can you find the unit rate and the slope

Solnce55 [7]

To find slope use the formula M = y2 - y1
-----------
x2 - y1
say you were given this (2,3) (-4,5) 2 would be x1 and 3 would be y1, -4 would be x2 and 5 would be y2 then just put it into the formula 5 - 3
--------
-4 - 2
which would be 2/ -6 which would simplify to 1/3

5 0

4 years ago

What is the y-intercept of the function f(x)=4 - 5x? 05 O 5

lapo4ka [179]

Answer:

Step-by-step explanation:

7 0

2 years ago

Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. (a) Descr

Daniel [21]

Answer:

(a) The sampling distribution of $\overline{X}$ = Population mean = 79

(b) P ( $\overline{X}$ greater than 81.2 ) = 0.1357

(c) P ( $\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of $\overline{X}$ = Population mean = 79

(b) What is Upper P ( $\overline{X}$ greater than 81.2 ) =

$P(\overline{X}> 81.2)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

= $P(Z> 1.1)$

= $1 - P(Z< 1.1)$

= 1 - .8643 =

= 0.1357

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

= $P(Z\leq -2.3)$

= .0107

(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$