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3 years ago

In surveying families with pets, 5 out of 7 families owned a dog. If 35 families had dogs, how many families were surveyed?

1 answer:

5 0

Hello There!

Follow these steps:
35/5 = 7
7 x 7 = 49
35 out of 49 families owned a dog.
49 families were surveyed.

Hope This Helps You!
Good Luck :)

- Hannah ❤

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Type the correct answer in each box. A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equatio

Galina-37 [17]

Answer:

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

Step-by-step explanation:

Given:

Center of circle is at (5, -4).

A point on the circle is $(x_1,y_1)=(-3, 2)$

Equation of a circle with center $(h,k)$ and radius 'r' is given as:

$(x-h)^2+(y-k)^2=r^2$

Here, $(h,k)=(5,-4)$

Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:

$r^2=(h-x_1)^2+(k-y_1)^2$

Using distance formula for the points (5, -4) and (-3, 2), we get

$r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100$

Therefore, the equation of the circle is:

$(x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100$

Now, rewriting it in the form asked in the question, we get

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

4 0

3 years ago

According to a 2010 study conducted by the Toronto-based social media analytics firm Sysomos, 71% of all tweets get no reaction.

klemol [59]

Answer:

a) the expected number of tweets with no reaction is 71 tweets

b) the variance is 20.59 tweets² ( ≈ 21 tweets )and the standard deviation is

4.537 tweets (≈ 5 tweets)

Step-by-step explanation:

we can use the binomial probability distribution

P(x, N, p ) = N!/[(N-x)!x!] * p^x * (1-p)^(N-x)

where

x = number of successful events

N = population total

p = probability for success for every individual and independent event

P = probability for x successful events

in our case p = 71% = 71/100 (probability of a tweet without reaction) , N = 100 tweets

a) the expected number of tweets

E(x) = N* p = 100 tweets * 71/100 = 71 tweets

b) the variance is

V(x) = N * p * (1-p)

V = 100 * 71/100 * (1-71/100) = 71* 29/100 = 20.59 tweets² ( units of variance are [N²] )

the standard deviation is

s = √V = √20.59 tweets² = 4.537 tweets

6 0

2 years ago

The product of a number and -3, increased by 8, is 20

natita [175]

answer:

-4

Step-by-step explanation:

-3x+8= 20

-8 -8

-3x = 12

÷-3. = ÷-3

x = -4

3 0

2 years ago

Simplify the expression: 10x + -8x + -10x+ -10 + 8x + -6x

Lynna [10]

Quilt

Step-by-step explanation:

8 0

2 years ago

Radioactive Decay:

Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

Answer: The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

Step-by-step explanation:

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}$

The formula used to calculate the time period for a first order reaction follows:

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}$

where,

k = rate constant = $0.01506days^{-1}$

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

$t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days$

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

7 0

2 years ago