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Ann [662]
2 years ago
12

The average daily rainfall for the past week in the town of Hope Cove is normally distributed, with a mean rainfall of 2.1 inche

s and a standard deviation of 0.2 inches. If the distribution is normal, what percent of data lies between 1.9 inches and 2.3 inches of rainfall? a) 95% b) 99.7% c) 34% d) 68%
Mathematics
1 answer:
natita [175]2 years ago
6 0

Answer:

D

Step-by-step explanation:

We calculate the z-score for each

Mathematically;

z-score = (x-mean)/SD

z1 = (1.9-2.1)/0.2 = -1

z2 = (2.3-2.1)/0.2 = 1

So the proportion we want to calculate is;

P(-1<x<1)

We use the standard score table for this ;

P(-1<x<1) = P(x<1) -P(x<-1) = 0.68269 which is approximately 68%

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A certain shade of pink is created by by adding 3 cups of red paint to 7 cups of white paint. How many cups of red paint should
Nonamiya [84]

Answer:

→[ Certain shade=red paint+ white paint]

→3 cups of red paint=7 cups of white paint

Dividing both side by 7,we get

→ 1 cup of white paint=3/7 cups of white paint

So, 3/7 of a cup should be added to 1 cup of white paint to make certain kind of shape.

Cups of white paint: Cups of red paint= 1. : \frac{7}{3}=\frac{3}{7}

k=\frac{3}{7} is constant of proportionality.



8 0
2 years ago
Read 2 more answers
Look at the figure. Which step should be taken next to construct a line through point P perpendicular to BA?
alexira [117]

ANSWER:

C. Place the compass on point A. Open the compass to a point between point P and point B.

EXPLANATION:

A perpendicular is a line that would be at a right angle to line BA.

The next step is to chose a radius that is greater than PB or PA so as to construct the bisector. And this can be done by placing the compass on point A, and open the compass to a point between point P and point B.

Use this radius to draw an arc above and below the line, and repeat the same using B as the center with the same radius. This would form two intersecting arcs above and below line BA. Join the point of intersection of the arcs by a straight line through P. This is the bisector of line BA through point P.

8 0
2 years ago
The function f(x) = 5One-fifth is reflected over the y-axis. Which equations represent the reflected function? Select two option
postnew [5]

Answer:

The equations that represent the reflected function are

f(x)=5(\frac{1}{5})^{-x}

f(x)=5(5)^{x}

Step-by-step explanation:

The correct question in the attached figure

we have the function

f(x)=5(\frac{1}{5})^{x}

we know that

A reflection across the y-axis interchanges positive x-values with negative x-values, swapping x and −x.

therefore

f(−x) = f(x).

The reflection of the given function across the y-axis will be equal to

(Remember interchanges positive x-values with negative  x-values)

f(x)=5(\frac{1}{5})^{-x}

An equivalent form will be

f(x)=5(\frac{1}{5})^{(-1)(x)}=5[(\frac{1}{5})^{-1})]^{x}=5(5)^{x}

therefore

The equations that represent the reflected function are

f(x)=5(\frac{1}{5})^{-x}

f(x)=5(5)^{x}

9 0
2 years ago
Read 2 more answers
In the diagram shown of circle A, segments UV and UT are congruent. If
grandymaker [24]

Answer:

 35°

Step-by-step explanation:

We assume that points S, T, U, V all lie on the circle.

Inscribed angle TUV intercepts long arc VST and short arc VUT. The long arc has measure 220°, so the measure of the short arc is 360° -220° = 140°.

Since chords UV and UT are the same length, point U bisects arc VUT. That means the measure of arc VU is 140°/2 = 70°.

Inscribed angle VSU intercepts arc VU, so has half the measure of the arc:

 ∠VSU = 70°/2

 ∠VSU = 35°

Read more on Brainly.com - brainly.com/question/15559971#readmore

8 0
2 years ago
Segment GI is congruent to Segment JL and Segment GH is congruent to Segment KL. I have to prove Segment HI is congruent to Segm
madreJ [45]

Answer:

See explanation

Step-by-step explanation:

1 step: \overline{GI}\cong \overline {JL} - given

2 step: \overline{GI}\cong \overline{GH}+\overline{HI} - Segments Addition Postulate

3 step: \overline{GH}+\overline{HI}\cong \overline {JL} - Substitution Property

4 step: \overline {JL}\cong \overline {JK}+\overline {KL} - Segments Addition Postulate

5 step: \overline{GH}+\overline{HI}\cong \overline {JK}+\overline {KL} - Substitution Property

6 step: \overline{GH}\cong \overline {KL} - given

7 step: \overline{GH}+\overline{HI}\cong \overline {JK}+\overline {GH} - Substitution Property of Equality

8 step: \overline{HI}\cong \overline {JK} - Subtraction Property of Equality

3 0
2 years ago
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