**Round to the nearest tenth Level ll: John missed 14 points on an 80 point science test. What percentage did he get?

Test 69,902 for divisibility by 2, 3, 5, 9, or 10.

Keith_Richards [23]

Its only divisible by 2.

5 0

3 years ago

Find the perimeter of the quadrilateral.

levacccp [35]

Question:

Find the perimeter of the quadrilateral. if x = 2 the perimeter is ___ inched.

The complete figure of this question is attached below

Answer:

<h3>The perimeter of the quadrilateral is 129 inches</h3>

Solution:

The complete figure of this question is attached below

Given that, a quadrilateral with,

Side lengths are:

$4x^2 + 8x\ inches \\\\3x^2-5x+20\ inches \\\\7x + 30\ inches \\\\31\ inches$

The values of the side lengths when x = 2 are

$(4x^2+8x)=(4\times 2^2+8\times 2)=(4\times 4+16)=16+16=32\ inch\\\\(3x^2-5x+20)=(3\times 2^2-5\times 2+20)=(3\times 4-10+20)=12+10=22\ inch\\\\(7x+30)=(7\times 2+30)=14+30=44\ inch$

Perimeter of a quadrilateral = Sum of its sides

Perimeter of given quadrilateral = 32 + 22 + 44 + 31 = 129 inches

Thus perimeter of the quadrilateral is 129 inches

5 0

3 years ago

EXPERTSSSS HELP (q.5)

Ostrovityanka [42]

Answer:

not paying 20% means that she DID pay %80

Step-by-step explanation:

4 0

3 years ago

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp

motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305$