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Anarel [89]
3 years ago
5

Please Help.... 50 points.

Mathematics
2 answers:
Aleksandr [31]3 years ago
5 0

Answer:

94%

Step-by-step explanation:

We need to look at the graph. Olive wants the probability that <em>20 or more </em>of the 50 babies were female. Looking at the graph, the x-axis is "number of females born out of 50 babies". We want to find the sum of all the ones that are at least 20, or 20 and above.

The numbers above the bars represents how many of each case there are. Above the bar labelled 20, there is the number 11, and so on. So, our sum is:

11 + 17 + 14 + 23 + 16 + 29 + 16 + 23 + 14 + 6 + 9 + 3 + 1 + 2 + 1 + 1 + 1 = 187

In total, there are 200 cases, so our probability is 187 / 200 = 93.5% ≈ 94%.

The answer is thus A.

iragen [17]3 years ago
3 0

Step-by-step explanation:

<u>Step 1:  Count up how many babies are more than 20</u>

<u />11 + 17 + 14 + 23+16+29+16+23+14+6+9+3+1+2+1+1+1

187

187 / 200 ← Will give us percentage of the probability that 20 or more of the 50 babies were born female.

0.935 * 100

93.5%

Answer:  Option A, 94%

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A carnival tent is set up using a 32-foot pole as the center. The Support wires each have an angle of elevation of 38 degrees fr
BARSIC [14]

Answer:

about 52 feet

Step-by-step explanation:

The figure is omitted--you may draw it to confirm my answer.

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2 years ago
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4 years ago
Read 2 more answers
Sean b= 5,79; c= 10,4,el angulo A= 54,46°, el angulo C mide ?
Rzqust [24]

Answer:

C \approx 91.732^{\circ}

Step-by-step explanation:

(This exercise is presented in Spanish and for that reason explanation will be held in such language)

El lado restante se determina por la Ley del Coseno:

a = \sqrt{b^{2}+c^{2}-2\cdot b\cdot c \cdot \cos A}

a = \sqrt{5.79^{2}+10.4^{2}-2\cdot (5.79)\cdot (10.4)\cdot \cos 54.46^{\circ}}

a \approx 8.466

Finalmente, el angulo C se halla por medio de la misma ley:

\cos C = - \frac{c^{2}-a^{2}-b^{2}}{2\cdot a \cdot b}

\cos C = -\frac{10.4^{2}-8.466^{2}-5.79^{2}}{2\cdot (8.466)\cdot (5.79)}

\cos C = -0.030

C \approx 91.732^{\circ}

3 0
3 years ago
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