1 - 4 + 6^2 ÷ ( 1 + 2 × 2 )

Lanie’s room is in the shape of a parallelogram.

VladimirAG [237]

Answer:yes 108 is more than 60

Step-by-step explanation:

6 0

2 years ago

Point Fis on line segment EG. EG = 100, EF = 4x - 20, and FG = 2x + 30. Write an equation that you would use to solve for x, the

andrew-mc [135]

Step-by-step explanation:

Line segment = EG

point on a line segment = F

EG = 100

EF = 4x - 20

FG = 2x + 30

a/q, EF + FG = EG

the equation is

4x - 20 + 2x + 30 = 100

now solution for x is

→ 4x - 20 + 2x + 30 = 100

→ 6x + 10 = 100

→ 6x = 100 - 10 = 90

→ x = 90/6 = 15

therefore, value of x is 15.

and according to given information

EF = 4x - 20 = 4(15) - 20 = 60 - 20 = 40

FG = 2x + 30 = 2(15) + 30 = 30 + 30 = 60

hope this answer helps you dear..take care and may u have a great day ahead!

3 0

3 years ago

Find the angle of least nonnegative measure, θC, that is coterminal with θ=<br> 9π/4.

Lostsunrise [7]

9514 1404 393

Answer:

θC = π/4

Step-by-step explanation:

To find the desired angle, subtract multiplies of 2π until the angle is in the desired range.

9π/4 -2π = (9-8)π/4 = π/4

The angle θC = π/4 is coterminal with θ = 9π/4.

8 0

2 years ago

A cat eats 2/3 cup of food each day. How many days will a bag containing 16<br> cuns of food last?

Masja [62]

Answer:

The correct answer is 24

7 0

3 years ago

A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar

Arisa [49]

Answer:

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

$n=17$ represent the sample selected

$\alpha=0.1$ represent the significance

$s^2 =4.7^2 =22.09$ represent the sample variance

$\sigma^2_0 =3.5^2 =12.15$ represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 12.15$

Alternative hypothesis: $\sigma^2 \neq 12.15$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom are:

$df =n-1= 17-1=16$

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0

3 years ago

1 - 4 + 6^2 ÷ ( 1 + 2 × 2 )​

1 - 4 + 6^2 ÷ ( 1 + 2 × 2 )