Help please! anything is appreciated, giving brainliested:)

A box of 42 jelly beans contains 9 orange jelly beans. What is the probability that a randomly selected jelly bean will be orang

umka2103 [35]

Answer: 3/14

Step-by-step explanation: P(Orange jelly bean) = 9/42 = 3/14

7 0

3 years ago

A system of equations has infinitely many solutions. If 2y – 4x = 6 is one of the equations, which could be the other equation?

Molodets [167]

Hi there!

$\large\boxed{y = 2x + 6}$

For a system to have infinite solutions, the lines have to be the same. We can begin by rearranging the given equation into the format y = mx + b:

2y - 4x = 6

Move x to the opposite side:

2y = 4x + 6

Divide all terms by 2:

y = 2x + 6

The answer choice that matches this is the first one.

5 0

3 years ago

A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed

tresset_1 [31]

Answer:

a) $z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) $P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

c) $z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

$X \sim N(\mu = 34, \sigma=2.5)$

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

$P(34 < X$

We can find the number of deviation from the mean with the z score formula:

$z= \frac{X -\mu}{\sigma}$

And replacing we got

$z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

$P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

$P(29 < X$

And replacing we got:

$z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

7 0

3 years ago

A rectangular prism has a length of 5 cm, a width of 2 1/2 cm, and a expressions height of 1 1/2 cm. Which can be used to find t

Assoli18 [71]

To find the volume of a rectangular prism, we can use the formula [ V = l * w * h ]

V = 5 * 2 1/2 * 1 1/2

V = 12 1/2 * 1 1/2

V = 18 3/4cm³

Now that we know the volume, we can check each answer choice.

Option A:

5 ∙ 2 1/2 ∙ 1 1/2

12 1/2 * 1 1/2

18 3/4cm³

Option B:

12 1/2 ∙ 1 1/2

18 3/4cm³

Option C:

1 1/2(5 + 2 1/2)

1 1/2(7 1/2)

11 1/4cm³

Option D:

1/2(12 1/2 ∙ 1 1/2)

1/2(18 3/4)

9 3/8cm³

Therefore, the only correct options are A and B.

Best of Luck!

5 0

4 years ago

Read 2 more answers

2.The mean area of several thousand apartments in a new development is advertised to be 1,100 square feet. A consumer advocate h

KiRa [710]

Answer:

Null Hypothesis, $H_0$ : $\mu$ = 1,100 square feet

Alternate Hypothesis, $H_A$ : $\mu$ < 1,100 square feet

Step-by-step explanation:

We are given that the mean area of several thousand apartments in a new development is advertised to be 1,100 square feet.

A consumer advocate has received numerous complaints that the apartments are smaller than advertised.

Let $\mu$ = mean area of several apartments.

So, Null Hypothesis, $H_0$ : $\mu$ = 1,100 square feet

Alternate Hypothesis, $H_A$ : $\mu$ < 1,100 square feet

Here, null hypothesis states that the mean area of apartments are same as advertised.

On the other hand, alternate hypothesis states that the mean area of apartments are smaller than advertised.

So, this would be the appropriate null and the alternative hypothesis to test this claim.

4 0

3 years ago