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3 years ago

can someone please help me?!

Mathematics

2 answers:

ololo11 [35]3 years ago

7 0

X = 19
Z = 28
Hope this helps

Aloiza [94]3 years ago

3 0

Answer:

x = 119

z = 28

Step-by-step explanation:

To find x: Because of same-side exterior angles, you can add x and 61 together to find 180 degrees.

x + 61 = 180

Subtraction: x = 119

To find z: Use given quantities, not found variables for most accurate answer. Because of opposite exterior angles, you can equal them to each other.

61 = 5z - 79

Addition: 140 = 5z

Division: 140/5 = 5z/5

Final Answer = 28 = z

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Alright , help me out please (10 points)

Ksju [112]

Answer:

the first one

the third one

the fourth one

Step-by-step explanation:

1. x does equal 9

2. the equation would lead to x being canceled out so there would be an x in your answer

3. x does equal 30

4. x does equal 6

5. x should equal 135

7 0

3 years ago

Dennis is cutting construction paper into rectangles for a project he needs to cut one rectangle that is 12 inches times 15 1/4

Nataly_w [17]

Answer:

Dennis need 288.91 square inches of construction paper for his project.

Step-by-step explanation:

Dennis is cutting construction paper into rectangles for a project.

He needs to cut one rectangle that is 12 inches times 15 1/4 inches

First convert the mixed fraction to improper fraction

$15\frac{1}{4} = \frac{(15\times4) + 1}{4} = \frac{61}{4}$

So the area of 1st rectangle is

$A_1 = 12\times \frac{61}{4} = 183 \: in^{2}$

He needs to cut another rectangle that is 10 1/3"" x 10 1/4""

The symbol " means inches

First convert the mixed fraction to improper fraction

$10\frac{1}{3} = \frac{(10\times3) + 1}{3} = \frac{31}{3}$

$10\frac{1}{4} = \frac{(10\times4) + 1}{4} = \frac{41}{4}$

So the area of 2nd rectangle is

$A_2 = \frac{31}{3}\times \frac{41}{4} = 105.91 \: in^{2}$

The total construction paper needed for this project is

$A = A_1 + A_2\\\\A = 183 + 105.91\\\\A = 288.91 \: in^{2}$

Therefore, Dennis need 288.91 square inches of construction paper for his project.

6 0

4 years ago

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

Blababa [14]

Let $X$ be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

$P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10$

where $Z$ follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

$P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)$

where $F_Z(z)=P(Z\le z)$ denotes the CDF of $Z$ , and $F_Z^{-1}$ denotes the inverse CDF. We have

$z_1=F_Z^{-1}(0.90)\approx1.2816$

Similarly, because 20% of students obtain less than 40 marks, we have

$P(X$

so that

$P(Z$

Then $\mu,\sigma$ are such that

$\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma$

$\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma$

and we find

$\mu\approx53.8739,\sigma\approx16.4848$

3 0

4 years ago

A group of people were asked if they had run a red light in the last year. 386 responded "yes", and 137 responded "no". Find the

ludmilkaskok [199]

Answer: The probability would 0.35770234986 or 35%

Step-by-step explanation:

3 0

3 years ago

F(x) = 4x - 9 i need helpp

Masteriza [31]

Answer:

Step-by-step explanation:

5 0

3 years ago

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￼can someone please help me?!

can someone please help me?!