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Wittaler [7]
2 years ago
11

7/8 - 5/12 ayudaaa por favor es para mañana​

Mathematics
1 answer:
RUDIKE [14]2 years ago
3 0
7/8 = 21/24
5/12 = 10/24

21/24-10/24 = 11/24
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when a spherical balloon is filled with air, it has a radius of 3 inches. what is the estimate for the volume of air in the ball
Crazy boy [7]

The volume of a sphere is given by the equation V=\frac{4}{3} \pi r^{3}, where V is the volume and r is the radius.


To answer this question, simply plug the correct values into the equation:

V=\frac{4}{3} \pi 3^{3} \\V=4.1888(27)\\V=113.0976


The volume of the air in the balloon is about 113.0976 in3.


Hope this helps!!

7 0
3 years ago
A box contains 10 tags, numbered 1 through 10, with a different number on each tag. A second box contains 8 tags, numbered 20 th
blagie [28]

Answer:

E) 29.0

Step-by-step explanation:

The value of the sum is obtained from 2 independent experiments: the value of the number of the first box X₁ and the value of the number of the second box X₂.

The expected value of a draw is the average of all its values, so E(X₁) = (1+2+3+4+5+6+7+8+9+10)/10 = 5.5 and E(X₂) = (20+21+22+23+24+25+26+27)/8 = 23.5

Hence, E(X₁+X₂) = E(X₁)+E(X₂) = 5.5+23.5=29

3 0
3 years ago
Solve the equation: <br><br> -4(m+6)=-8<br><br><br> m=-10<br><br> m=-4<br><br> m=3.5<br><br> m=26
Harrizon [31]
M = -4 for this one...............
8 0
2 years ago
Read 2 more answers
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0
3 years ago
Plz help plz plz plz I need help I would really appreciate if u help me out
allsm [11]
What’s the question tho
6 0
3 years ago
Read 2 more answers
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