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alex41 [277]
3 years ago
6

1) A population of bees is increasing at a rate of 8% per week. If

Mathematics
1 answer:
ad-work [718]3 years ago
3 0

Answer:

≈387,7

Step-by-step explanation:

I hope it's right. Sorry if it's not right.

You might be interested in
Write an equivalent expression to present patricia total earnings this week 7h + 135 + 5h = ?h + ?​
Paul [167]

Answer: 12 h + 135

Step-by-step explanation: 7h +135 + 5h = ?h + ?​

                                             7h +5h +135  

                                            =12h +135

So u combine like terms together  which are 7h and 5h which gives you 12 h.And then you have 135 left,so you add 12h to 135.

Ps.If you don't know what like terms are.I will explain down below.

Like terms are  variable that are the same.For an example 4y and 3y are like terms .Another example is 1/3 xy and 6xy .

Hope this helps :) and if it does please give brainliest .

8 0
2 years ago
Can anyone help me<br> Please and thankyou? Been struggling for 20minutes
schepotkina [342]

There is an app that will help for these types of questions called, "Showmath."

5 0
3 years ago
Read 2 more answers
-. A 6-foot-tall man is standing 50 feet from a flagpole. When he looks at the top of the flagpole,
taurus [48]

Answer:

The height of the flagpole is 46 feet

Step-by-step explanation:

First we find the ratio to connect the variables

Tan(∅) = opposite/adjacent

Plug in the Variables

Tan(39) = x/50

Solve for x

50*tan(39) = x

x = 40.48

Don't forget to include the man's height

40.48 + 6 = 46.48

7 0
3 years ago
Suppose there is a 13.9 % probability that a randomly selected person aged 40 years or older is a jogger. In​ addition, there is
Blababa [14]

Answer:

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

It would be unusual to randomly select a person aged 40 years or older who is male and jogs.

Step-by-step explanation:

We have these following probabilities.

A 13.9% probability that a randomly selected person aged 40 years or older is a jogger, so P(A) = 0.13.

In​ addition, there is a 15.6% probability that a randomly selected person aged 40 years or older is male comma given that he or she jogs. I am going to say that P(B) is the probability that is a male. P(B/A) is the probability that the person is a male, given that he/she jogs. So P(B/A) = 0.156

The Bayes theorem states that:

P(B/A) = \frac{P(A \cap B)}{P(A)}

In which P(A \cap B) is the probability that the person does both thigs, so, in this problem, the probability that a randomly selected person aged 40 years or older is male and jogs.

So

P(A \cap B) = P(A).P(B/A) = 0.156*0.139 = 0.217

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

A probability is unusual when it is smaller than 5%.

So it would be unusual to randomly select a person aged 40 years or older who is male and jogs.

4 0
3 years ago
Please help me!<br> Find the number x such that f(x) =1
STatiana [176]

Answer:

D

Step-by-step explanation:

We have the piecewise function:

f(x) = \left\{        \begin{array}{ll}            -\frac{1}{2}x-1 & \quad x \leq -2 \\            x & \quad x > -2        \end{array}    \right.

And we want to find x such that f(x)=1.

So, let's substitute 1 for f(x):

1 = \left\{        \begin{array}{ll}            -\frac{1}{2}x-1 & \quad x \leq -2 \\            x & \quad x > -2        \end{array}    \right.

This has two equations. So, we can separate them into two separate cases. Namely:

1=-\frac{1}{2}x-1\text{ or } 1=x

Let's solve for x in each case.

Case I:

We have:

1=-\frac{1}{2}x-1

Add 1 to both sides:

2=-\frac{1}{2}x

Let's cancel out the fraction by multiplying both sides by -2. So:

2(-2)=(-2)\frac{-1}{2}x

The right side cancels:

-4=x\\

Flip:

x=-4

So, x is -4.

Case II:

We have:

1=x

Flip:

x=1

This is the solution for our second case.

So, we have:

x_1=-4\text{ or } x_2=1

Now, can check to see if we have to to remove solution(s) that don't work.

Note that x=-4 is the solution to our first equation.

The first equation is defined only if x is less than -2.

-4 <em>is </em>less than -2. So, x=-4 is indeed a solution.

x=1 is the solution to our second equation.

The second equation is defined only if x is greater than or equal to -2.

1 <em>is</em> greater than or equal to -2. So, x=1 is <em>also </em>a solution.

Therefore, our two solutions are:

x_1=-4\text{ or } x_2=1

Out of our answer choices, we can pick D.

And we're done!

7 0
3 years ago
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