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2 years ago

13

A bat and a ball cost $1.10. The bat costs one dollar more than the ball. How much does the ball cost? Correct Answer Gets Brain

liest answer!

2 answers:

KatRina [158]2 years ago

8 0

Answer:

0.10

Step-by-step explanation:

If together the ball and bat cost $1.10 and the bat cost one more dollar than the ball than you would have too subtract $1 from what the ball and bat together. so 0.10 would be left

SpyIntel [72]2 years ago

5 0

The answer to that question would be 5c , I took it and got it right

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A coat originally cost $130. Now it is on sale for 60% of its original cost. What is the sale price of the coat?

babunello [35]

Answer:

$78

Step-by-step explanation:

to find 60% of 130 multiply 130 by 0.6

this gives you 78 dollars

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2 years ago

30PTs PLZ HelpASAP

Studentka2010 [4]

I think the answer is a

6 0

2 years ago

Read 2 more answers

HELP! What is AB? WILL GIVE BRAINLIEST!

Natalija [7]

AB = BC

3x-4=5x-10
-4=2x-10
6=2x
3=x

AB = 3x-4
AB = 3 (3)-4
AB = 9-4
AB =5

Therefore, AB is 5.

BC = 5x-10
BC = 5(3)-10
BC = 15-10
BC= 5

3 0

3 years ago

what is the probability of seeing a sample mean for 21 observations less or equal to the sample mean that we observed

kramer

Answer:

$P(x \le 21) = 0.69146$

Step-by-step explanation:

The missing parameters are:

$n = 64$ --- population

$\mu = 20$ --- population mean

$\sigma = 16$ -- population standard deviation

Required

$P(x \le 21)$

First, calculate the sample standard deviation

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{16}{\sqrt {64}}$

$\sigma_x = \frac{16}{8}$

$\sigma_x = 2$

Next, calculate the sample mean $\bar_x$

$\bar x = \mu$

So:

$\bar x = 20$

So, we have:

$\sigma_x = 2$

$\bar x = 20$

$x = 21$

Calculate the z score

$x = \frac{x - \mu}{\sigma}$

$x = \frac{21 - 20}{2}$

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

$x = 0.50$

So, we have:

$P(x \le 21) = P(z \le 0.50)$

From the z table

$P(z \le 0.50) = 0.69146$

So:

$P(x \le 21) = 0.69146$

4 0

2 years ago

Establish for which constants c the following function is a density.

Valentin [98]

By "density" I assume you mean "probability density function". For this to be the case for $f(x)$ , we require

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=1$

Since

$f(x)=\begin{cases}cx^{1/2}&\text{for }0$

you have

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_0^2cx^{1/2}\,\mathrm dx=\dfrac{2c}3x^{3/2}\bigg|_{x=0}^{x=2}=\dfrac{2^{5/2}c}3=1$

which means

$c=\dfrac3{2^{5/2}}=\dfrac3{4\sqrt2}$

7 0

2 years ago