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IceJOKER [234]
4 years ago
8

The diameter of a circular tray is 28cm. Find the area of the tray. Take[Pi=22/7] ​

Mathematics
1 answer:
adoni [48]4 years ago
8 0

Answer:

616 cm²

Step-by-step explanation:

<u>Area of circle:</u>

  • A = πr² = 1/4πd²

<u>Since d = 28 cm and π = 22/7, the area is:</u>

  • 1/4*22/7*28² = 616 cm²
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Work out, giving your answer in its simplest form: <br><br>3 1/2 ÷ 2 3/5​
Alecsey [184]

Answer:

this is the ans- 3.36956522

7 0
3 years ago
Please answer correctly !!!!!!!!!!!!!!! Will mark Brianliest answer !!!!!!!!!!!!!!!!!
ludmilkaskok [199]

Answer:

x = 8√2

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

<u>Trigonometry</u>

  • [Right Triangles Only]: Pythagorean Theorem: a² + b² = c²

Step-by-step explanation:

<u>Step 1: Define</u>

We are given a right triangle. We can use PT to solve for the missing length.

<u>Step 2: Identify Variables</u>

Leg <em>a</em> = 8

Leg <em>b</em> = 8

Hypotenuse <em>c</em> = <em>x</em>

<u>Step 3: Solve for </u><em><u>x</u></em>

  1. Substitute [PT]:                    8² + 8² = x²
  2. Exponents:                           64 + 64 = x²
  3. Add:                                      128 = x²
  4. Isolate <em>x</em>:                               8√2 = x
  5. Rewrite:                                x = 8√2
4 0
3 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
Pleasee help ASAP I’m begging u plssss
Colt1911 [192]
B, y=mx+b
b will always be the y intercept while is the slope
8 0
3 years ago
Question 31 pts Prove the statement is true using mathematical induction: 2n-1 ≤ n! Use the space below to write your answer. To
Sladkaya [172]

Answer:

P(3) is true since 2(3) - 1 = 5  < 3! = 6.  

Step-by-step explanation:

Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3

Basis: P(3) is true since 2(3) - 1 = 5  < 3! = 6.  

Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:

2(k+1) - 1 = 2k + 2 - 1

≤ 2 + k! (by the inductive hypothesis)

= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.

4 0
3 years ago
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