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2 years ago

Round 77.1110969489 to the nearest hundred-thousandth

Mathematics

1 answer:

Nezavi [6.7K]2 years ago

4 0

77.11110 it’s the nearest hundred- thousandth

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Juan bought n packs of pencils. Each pack has 15 Write an equation to represent the total number of pencils p that Juan bought.

pav-90 [236]

Answer:

15*n=p

Step-by-step explanation:

4 0

3 years ago

1. Flight 202's arrival time is normally distributed with a mean arrival time of 4:30

Bezzdna [24]

Answer:

Step-by-step explanation:

1) Let the random time variable, X = 45min; mean, ∪ = 30min; standard deviation, α = 15min

By comparing P(0 ≤ Z ≤ 30)

P(Z ≤ X - ∪/α) = P(Z ≤ 45 - 30/15) = P( Z ≤ 1)

Using Table

P(0 ≤ Z ≤ 1) = 0.3413

P(Z > 1) = (0.5 - 0.3413) = 0.1537

∴ P(Z > 45) = 0.1537

2) By compering (0 ≤ Z ≤ 15) ( that is 4:15pm)

P(Z ≤ 15 - 30/15) = P(Z ≤ -1)

Using Table

P(-1 ≤ Z ≤ 0) = 0.3413

P(Z < 1) = (0.5 - 0.3413) = 0.1587

∴ P(Z < 15) = 0.1587

3) By comparing P(0 ≤ Z ≤ 60) (that is for 5:00pm)

P(Z ≤ 60 - 30/15) = P(Z ≤ 2)

Using Table

P(0 ≤ Z ≤ 1) = 0.4772

P(Z > 1) = (0.5 - 0.4772) = 0.0228

∴ P(Z > 60) = 0.0228

6 0

3 years ago

4/5=/20 math...........

Mumz [18]

Answer:

1.) 16/20

2.) 15/35

3.) 12/9

4.) 52/60

5.) 9 inches

6.) 24 centimeters

7.) 16 logs

8.) $15.40

9.) 10 miles

10.) 1.5 inches

I hope this is good enough:

7 0

2 years ago

A rectangle has a length of (x+10) centimeters and a width of x centimeters. If the perimeter is 32 centimeters, what is the len

I am Lyosha [343]

Given:

Length of rectangle = (x+10) cm

Width of the rectangle = x cm

Perimeter = 32 cm

To find:

The length of the rectangle.

Solution:

We know that,

$Perimeter=2(l+w)$

Where, l is length and w is width.

Substituting the values, we get

$32=2((x+10)+x)$

$32=2(2x+10)$

$32=4x+20$

Subtract 20 from both sides.

$32-20=4x$

$12=4x$

Divide both sides by

$3=x$

So, the length is

$3+10=13$

Therefore, the length of the rectangle is 13 cm.

7 0

2 years ago

An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the

Sergio039 [100]

Answer:

The estimate is $P__{hat}} \pm E = 0.37 \pm 0.0348$

Step-by-step explanation:

From the question we are told that

The sample size is n = 522

The sample proportion of students would like to talk about school is $\r p__{hat}} = 0.37$

Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as

$\alpha = 100 - 90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645$

Generally the margin of error can be mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }$

=> $E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }$

=> $E = 0.0348$

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

$P__{hat}} \pm E$

substituting values

$0.37 \pm 0.0348$

5 0

3 years ago