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Jobisdone [24]
2 years ago
15

Solve this question please

Mathematics
1 answer:
Iteru [2.4K]2 years ago
4 0

The value of y in y = 5x if the value of x doubles itself is y also doubles itself; option A

<h3>How to solve equation?</h3>

y = 5x

if x = 2

y = 5(2)

y = 10

if the value of x doubles, for instance, if x = 4

y = 5(4)

y = 20

y = 5/x

if x = 0.5

y = 5/0.5

y = 10

if the value of x triples, for instance, if x = 0.125

y = 5/x

= 5/0.125

y = 40

The value of y in y = 5/x if the value of x triples itself is y also triples itself; option B

Learn more about equation:

brainly.com/question/13763238

#SPJ1

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6m + 2 &lt; 5m-4 solve the inequality for m
djyliett [7]

Answer:

m < -6

Step-by-step explanation:

plz can u mark brainliest if correct. thank you so much.

6 0
3 years ago
Please help me out!!!!!!!
maksim [4K]

Answer: 1%

Explanation:

Add up the frequencies in the bottom row: 350+200+245+125+66+10+4 = 1000

There are 1000 families total. Of this total, 10+4 = 14 families have more than six people. I added up the frequency counts for "7 people" and "8 people" to get 14.

Divide 14 over 1000 and we get 14/1000 = 0.014 = 1.4% which rounds to 1%

7 0
3 years ago
The amount of time it takes a bat to eat a frog was recorded for each bat in a random sample of 12 bats. The resulting sample me
spin [16.1K]

Answer: a. CI for the mean: 17.327 < μ < 26.473

b. CI for variance: 29.7532 ≤ \sigma^{2} ≤ 170.9093

Step-by-step explanation:

a. To construct a 95% confidence interval for the mean:

The given data are:

mean = 21.9

s = 7.7

n = 12

df = 12 - 1 = 11

1 - α = 0.05

\frac{\alpha}{2} = 0.025

t-score = t_{0.025,11} = 2.2001

Note: since the sample population is less than 30, it is used a t-score.

The formula for interval:

mean ± t.\frac{s}{\sqrt{n} }

Substituing values:

21.9 ± 2.200.\frac{7.7}{\sqrt{12} }

21.9 ± 4.573

The interval is: 17.327 < μ < 26.473

b. A 95% confidence interval for the variance:

The given values are:

s^{2} = 7.7^{2}

s^{2} = 59.29

α = 0.05

\frac{\alpha}{2} = 0.025

1-\frac{\alpha}{2} = 0.975

\chi^{2}_{0.025,11} = 21.92

\chi^{2}_{0.975,11} = 3.816

Note: To find the values for \chi^{2}_{\alpha/2,n-1} and \chi^{2}_{1-\alpha/2,n-1}, look for them at the chi-square table

The formula to calculate interval:

(\frac{(n-1).s^{2}}{\chi^{2}_{\alpha/2,n-1}} , \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2,n-1}})

are the lower and upper limits, respectively.

Substituing values:

(\frac{11.59.29}{21.92} , \frac{11.59.29}{3.816})

(29.7532, 170.9093)

The interval for variance is: 29.7532 ≤ \sigma^{2} ≤ 170.9093

6 0
3 years ago
Solve the system of equations. y = –4x y = 2x^2 – 15x
dexar [7]

Answer:

X=0, x= 11/2. y=0, y=-22

7 0
3 years ago
PLEASE HELP ASAP!!!! I GIVE A BRAINLIEST TO THE RIGHT ANSWER!!!!
Mandarinka [93]
Use cosine law to solve the problem. See image attached.
What we're looking for is the length of side A (a side which is opposite with angle A).

This is the formula of cosine law
A² = B² + C² - 2BC cos A

Input the numbers
A² = 75² + 90² - 2(75)(90) cos 85°
A = \sqrt{75^{2}+90^{2}-2(75)(90)cos85^{0}}
A = \sqrt{90^{2}+75^{2}-2(90)(75)cos85^{0}}

4 0
3 years ago
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