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3 years ago

How many tens and ones in the number 82?

2 answers:

6 0

The answer is 8 tens, 2 ones. The 8 is in the tens place and the 2 is in the ones place. Look up place value, my friend!

Elina [12.6K]3 years ago

4 0

8 tens and 2 ones......

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The circumference of the circular table on Beverly’s porch is 72pi inches. What is the radius of the table

sergejj [24]

Your answer is 18 :)

work:
pi(2)d
pi(72)
72/2=36=d

2r=d
36/2=18=r

6 0

3 years ago

Read 2 more answers

Find the value of x

avanturin [10]

Answer:

Step-by-step explanation:

1. 56 + 81 = 137

2. The exterior angle theorem means that the 2 angles within a triangle that is on the opposite side of the angle outside the triangle could be added to find the angle that is outside of the triangle.

3. 56 and 81 are on the opposite side of the "x" outside the triangle which means that they could be added together which equals 137 degrees.

Hope this helps!

3 0

3 years ago

Read 2 more answers

Which ordered pair is a solution of the equation? 2x+4y=6x-y OPTION A: Only (4,5) OPTION B: Only (5,4) OPTION C Both A and B OPT

sammy [17]

Answer:

option b

Step-by-step explanation:

replace x and y with the x and y of the ordered pair

option a: 2(4)+4(5)=6(4)-5

solve

8+20=24-5

28=19 not true

option b:2(5)+4(4)=6(5)-4

solve

10+16=30-4

26=26 true

6 0

3 years ago

A calculator is broken so that the only keys that still work are the $\sin$, $\cos$, $\tan$, $\cot$, $\arcsin$, $\arccos$, and $

miskamm [114]

What is this check the q again

7 0

2 years ago

A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determinin

riadik2000 [5.3K]

Answer:

Option b - not significantly greater than 75%.

Step-by-step explanation:

A random sample of 100 people was taken i.e. n=100

Eighty of the people in the sample favored Candidate i.e. x=80

We have used single sample proportion test,

$p=\frac{x}{n}$

$p=\frac{80}{100}$

$p=0.8$

Now we define hypothesis,

Null hypothesis $H_0$ : candidate A is significantly greater than 75%.

Alternative hypothesis $H_1$ : candidate A is not significantly greater than 75%.

Level of significance $\alpha=0.05$

Applying test statistic Z -proportion,

$Z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

Where, $\widehat{p}=80\%=0.80$ and $p=75%=0.75$

Substitute the values,

$Z=\frac{0.80-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}$

$Z=\frac{0.80-0.75}{\sqrt{\frac{0.1875}{100}}}$

$Z=\frac{0.05}{0.0433}$

$Z=1.1547$

The p-value is

$P(Z>1.1547)=1-P(Z$

$P(Z>1.1547)=1-0.8789$

$P(Z>1.1547)=0.1241$

Now, the p-value is greater than the 0.05.

So we fail to reject the null hypothesis and conclude that the A is not significantly greater than 75%.

Therefore, Option b is correct.

7 0

3 years ago