What is the answer to number 4 about the lake?

A store sold 200 cell phone covers in April. The circle graph shows the

Brilliant_brown [7]

Answer:

24

Step-by-step explanation:

Given

Total is 200

From the attachment, we have:

Required

Black- 35%

Blue-23%

How many more of black than blue were sold

First, calculate the difference in percentage sold

Multiply by the total phones sold;

Hence, 24 more blacks were sold than blue

Hope this helps

Brainliest appreciated :)

6 0

2 years ago

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The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0

3 years ago

Sally plans to enclose a square garden with fencing. The area of the garden is 1600 square feet. How many feet of fencing does s

Llana [10]

Answer:

160 ft

Step-by-step explanation:

the length of one side would be √1600 which is 40 so 40• 4 is 160

4 0

3 years ago

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Can someone help me with this

antiseptic1488 [7]

Your answer is A

24.6

5 0

3 years ago

You design a new app for cell phones. Your revenue for x downloads is given by f(x) = 3x Your profit

Snowcat [4.5K]

Answer:

p = 2.7x - 20

p(90) = $223

Step-by-step explanation:

revenue → f(x) = 3x ,where x is the number of downloads.

the statement “Your profit p is $20 less than 90% of the revenue for x downloads”

means

$p=90\% f\left( x\right) -20$

means

$p=\frac{90}{100} \times f\left( x\right) -20$

means

$p=\frac{9}{10} \times (3x) -20$

means

$p=\frac{9}{10} \times (3x) -20$

means

$p=2.7x -20$

………………………………

the profit for 90 downloads :

= p(90)

= 2.7×(90) - 20

= $223

5 0

1 year ago