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Oksana_A [137]
3 years ago
5

To prepare for a bake-off, Lee used 640 grams of flour each day for 4 weeks. How many kilograms did Lee use in those 4 weeks?

Mathematics
1 answer:
dangina [55]3 years ago
8 0
1week = 7days so
4weeks = 28 days
So 28 x 640 = 17920

1000grams = 1kg

17920 divide 1000 =17.92

Lee uses 17.92kg in those 4weeks
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Step-by-step explanation:

such a scale factor applies to all sides and lines.

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12×f = 4

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3 years ago
In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311
Bingel [31]

Answer:

Step-by-step explanation:

Hello!

The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:

X₁: Number of Asians that would welcome a white person into their families.

X₂: Number of Asians that would welcome a Latino person into their families.

X₃: Number of Asians that would welcome a black person into their families.

Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251

Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.

Z_{1-\alpha /2}= Z_{0.975}= 1.965

1. 95% CI for Asians that would welcome a white person.

If 79% would welcome a white person, then the expected value is:

E(X)= n*p= 251*0.79= 198.29

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409

√V(X)= 6.45

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

198.29±1.965*6.45

[185.62;210.96]

With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.

2. 95% CI for Asians that would welcome a Latino person.

If 71% would welcome a Latino person, then the expected value is:

E(X)= n*p= 251*0.71= 178.21

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809

√V(X)= 7.19

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

178.21±1.965*7.19

[164.08; 192.34]

With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.

3. 95% CI for Asians that would welcome a Black person.

If 66% would welcome a Black person, then the expected value is:

E(X)= n*p= 251*0.66= 165.66

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244

√V(X)= 7.50

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

165.66±1.965*7.50

[150.92; 180.40]

With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.

I hope it helps!

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A. & B. Is the correct answer.

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HELP!!!!! ASAP PLEASE!!!!!!!!!!
Nataly_w [17]

Answer:

Tangent line states that a line in the plane of a circle that intersect the circle in exactly one point.

Common external tangent states that a common tangent that does not intersects the line segment joining the centers of circle.

Common internal tangent states that a common tangent that intersects the line segment joining the centers of circle.

Circumscribe polygon states that a polygon with all sides tangent to a circle contained within the polygon.

Therefore:

A polygon with all sides tangent to a circle contained within the polygon = Circumscribe polygon

A common tangent that intersects the line segment joining the centers of circle = Common internal tangent

A common tangent that does not intersects the line segment joining the centers of circle = Common external tangent

a line in the plane of a circle that intersect the circle in exactly one point = Tangent line

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