Can someone help please

The swim camp cost $280. How long will it take Kareem to save enough for tbe camp?

aev [14]

I want to help you but your question is missing multiple variables. To answer the question you need to know how much Kareem is making. Whether its a certain amount per task or per day or over a certain amount of time. And it depends on how long she has to save up for camp you know when camp starts and if she has enough time to have enough money saved up. You can message me if you have anymore questions about this. Hope this was helpful.

6 0

4 years ago

the temperature t at which water boils is inversely proportional to the number of feet F the water is above sea level. write the

Fudgin [204]

Answer:

Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula

T

=

14

,

000

d

gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.

3 0

3 years ago

Find the derivative of <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl

aev [14]

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$

8 0

3 years ago

Adelaide paid $35 for a manicure and gave the manicurist a 20% tip.

Nikolay [14]

That would be 35 + 0.20*35 = $42

7 0

3 years ago

A Niffler is a magical creature who loves to collect shiny objects and store them in their pouch, much like a kangaroo. The amou

Fudgin [204]

Answer:

Probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is 0.1515.

Step-by-step explanation:

We are given that the amount a Niffler can hold in their pouch is approximately normally distributed with a mean of 25 pounds of shiny objects and a standard deviation of 6.8 pounds.

Let X = amount a Niffler can hold in their pouch

So, X ~ Normal( $\mu=25,\sigma^{2} =6.8^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 25 pounds

$\sigma$ = standard deviation = 6.8 pounds

Now, the probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is given by = P(X > 32 pounds)

P(X > 32 pounds) = P( $\frac{X-\mu}{\sigma}$ > $\frac{32-25}{6.8}$ ) = P(Z > 1.03) = 1 - P(Z $\leq$ 1.03)

= 1 - 0.8485 = 0.1515

The above probability is calculated by looking at the value of x = 1.03 in the z table which has an area of 0.8485.

Hence, the probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is 0.1515.

4 0

3 years ago