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Sholpan [36]
3 years ago
7

Which of the bar graphs could be used to represent the data shown by the histogram

Mathematics
1 answer:
serg [7]3 years ago
4 0

Answer:

B (red)

Step-by-step explanation:

This is a unimodal histogram since there is only one data set that is most represented. Meaning that the red data set would represent the entire histogram since it's the majority.

Hopefully this helps.

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Gabriel makes a model of a pyramid with the dimensions shown.
professor190 [17]

Answer:

The Area of the square base is 114 In .2

The Area of each triangle face is 66 in .2

Gabriel will need 408 in .2 of paint

6 0
3 years ago
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The Width of a rectangular Park is 10 metres shorter than its length.if the area of the park is 236 square metre find its length
madam [21]

Answer:

let length be l

breadth=l-10m

area of rectangular park=236m²

Step-by-step explanation:

area of rectangular park=236m²

l×(l-10)=236

l²-10l-236=0

l²-2.l.5+5²-236-5²=0

(l-5)²-236-25=0

(l-5)²=261

(l-5)=√261

(l-5)=16.15

l=16.15+5=21.15

b=l-10=21.15-10=11.15

3 0
2 years ago
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Write in if/Then form: A shape has 4 congruent sides, if it is a rhombus.
8_murik_8 [283]

Answer: If a shape has 4 congruent sides, then it is a rhombus.

Step-by-step explanation: I just added the words in the appropriate places.

8 0
3 years ago
Area of a rectangle 8cm long and 4cm in height. please tell me how to do this also, I'm so confused.
Tema [17]
The area of a rectangle is easy to compute.
Area = length times height
Area = 8 cm * 4 cm = 32 square centimeters
6 0
3 years ago
All bags entering a research facility are screened. The screening process is not perfect so that 77% of the bags that contain fo
garik1379 [7]

Answer:

(1) 0.5933

(2) 0.8955

Step-by-step explanation:

We are given that all bags entering a research facility are screened.

Let Probability that bags entering the building contains forbidden material,

 P(F) = 0.69

Probability that bags entering the building does not contains forbidden material,   P(NF) = 1 - 0.69 = 0.31

Let event A = alarm gets triggered

Probability that alarm gets trigger given the bags contain forbidden material, P(A/F) = 0.77

Probability that alarm gets trigger given the bags does not contain forbidden material, P(A/NF) = 0.20

(1) Probability that a bag triggers the alarm, P(A) ;

         P(A) = P(F) * P(A/F) + P(NF) * P(A/NF)

                 = (0.69 * 0.77) + (0.31 * 0.20) = 0.5313 + 0.062

                 = 0.5933

Therefore, probability that a bag triggers the alarm is 0.5933 .

(2) Probability that a bag that triggers the alarm will actually contain forbidden material is given by P(F/A) ;

Using Bayes' Theorem;

    P(F/A) = \frac{P(F) * P(A/F)}{P(F) * P(A/F) + P(NF) *P(A/NF)} = \frac{0.69*0.77}{0.69*0.77+0.31*0.20} = \frac{0.5313}{0.5933}

               = 0.8955

6 0
3 years ago
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