How expanded form 7915

Twenty-five less two times a number, x, is fifteen.

Karo-lina-s [1.5K]

Answer:

x=20

Step-by-step explanation:

2x-25=15

2x=40

x=20

8 0

3 years ago

it costs 5.50 per hour to rent a pair of ice skates, for up to 2 hours. after 2 hours, the rental cost per hour decreases to 2.5

Temka [501]

2 x 5.50=11. 2x2.50=5 11+5=16. Answer $16

7 0

3 years ago

A man on a 135-ft vertical cliff looks down at an angle of 16°13' and sees his friend.

iren2701 [21]

A) Use the trigonometric function and ratio cos(x) = A/H. (Adjacent/Hypotenuse)
1. cos(x) = 135/H
2. H*cos(16°) = 135
3. H = 135/cos(16°)
***Make sure calculator is in degree mode***
4. H = 140.44 ft (rounded to two decimal places)
________

b) Use the function and ratio tan(x) = O/A (Opposite/Adjacent)
1. tan(x) = O/A
2. tan(16°) = O/135
3. 135*tan(16°) = O
4. O = 38.71 ft (two decimal places)

Good luck!

6 0

3 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

(c) What is Upper P ( $\overline{X}$ less than or equals 74.4 ) =

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

= $P(Z\leq -2.3)$

= .0107

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

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