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3 years ago

13+(4 to the 3rd power ÷2 )×5-7

Mathematics

2 answers:

frozen [14]3 years ago

7 0

Answer:

166

Step-by-step explanation:

PEMDAS

1 () 2 Exponents 3 Multiplication 4 Divide 5 Addition 5 Subtraction

4 to the power of 3 is 64

64 ÷ 2 is 32

32 × 5 is 160

160 + 13 is 173

173 - 7 is 166

Vinvika [58]3 years ago

3 0

Answer:185

Step-by-step explanation:

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What is the standard form of y = 6(x - 2)2 - 3?

AveGali [126]

Answer I think A

Step-by-step explanation:

4 0

3 years ago

A team t-shirt costs $3 per adult and $2 per child. On a certain day, the total number of adults (a) and children (c) who bought

lesya [120]

A+c=100

3a+2c=275, from the first c=100-a making the 2nd equation become:

3a+2(100-a)=275 perform indicated multiplication on left side

3a+200-2a=275 combine like terms on left side

a+200=275, subtract 200 from both sides

a=75, and since c=100-a

c=100=75=25

So the answer is D. 25 children and 75 adults
Equation 1: a+c=100
Equation 2: 3a+2c=275

8 0

3 years ago

3. Find the values of the variables and the lengths of the sides of this kite.

Kazeer [188]

Answer:

So, the correct option is A. x = 9, y = 14 , Sides are 11 and 20

Step-by-step explanation:

P.S - the exact question is -

As given in the question ,

Side AB = Side AD

⇒y - 3 = x + 2

⇒y - x = 2 + 3 = 5

⇒y - x = 5 .........(1)

Also, given

Side BC = Side CD

⇒2x + 2 = x + 11

⇒2x - x = 11 - 2

⇒x = 9

∴ equation (1) becomes

y - 9 = 5

⇒y = 5 + 9 = 14

⇒y = 14

∴ we get

x = 9, y = 14

Now,

as given ,

Side AB = y - 3

⇒ AB = 14 - 3 = 11

Side BC = 2x + 2

⇒ BC = 2(9) + 2 = 18 + 2 = 20

Side CD = x + 11

⇒ CD = 9 + 11 = 20

Side DA = x + 2

⇒DA = 9 + 2 = 11

∴ we get

The sides are 11 and 20

So,

The correct option is A. x = 9, y = 14 , Sides are 11 and 20

8 0

2 years ago

Write the following number in expanded form.<br><br>176,945,596

Greeley [361]

100,000,000+70,000,000+6,000,000+ 900,000+40,000+5,000+500+90+6= 176,945,596
I hope this helped

5 0

2 years ago

Read 2 more answers

A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

7 0

3 years ago