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konstantin123 [22]

3 years ago

10

Jonah flipped a coin 10 times. He flipped heads 3 times and tails 7 times. What is the experimental probability that Jonah will

flip heads on his next flip?

1 answer:

steposvetlana [31]3 years ago

7 0

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ 3/10 = 0.3

The experimental probability would be 0.3

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Laura sold t-shirts at a festival. She made $4 for each t-shirt she sold. Her expenses were $25. If she made a profit of $71, ho

Mamont248 [21]

Answer:

24 t shirts

Step-by-step explanation:

revenue - expenses = profit

profit = $71

expenses = $25

revenue = money made from t shirts = $4 for each t shirt, x times (she makes $4 for each t shirt and sells x t shirts) = $4 * x

4 * x - 25 = 71

add 25 to both sides to isolate the variable and its coefficient

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Trisha is going to make a border of icing around a circular cake with a diameter of 8 inches which of the following represents t

nadya68 [22]

Answer:

B. 25.12

Step-by-step explanation:

We are simply finding the perimeter of the cake.

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Perimeter: 25.12 inches

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3 years ago

If y varies directly as x, and y = 25 when x = 5, find y when x = 25.

zimovet [89]

A.125
Because to get from 5 to 25 you need to multiply by 5 so 25 times five equals 125

6 0

3 years ago

4. Six numbers are shown.

Reptile [31]

Step-by-step explanation:

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7 0

3 years ago

Use technology or a z-score table to answer the question.

Alik [6]

Answer:

The second choice: Approximately $65.2\%$ of the pretzel bags here will contain between 225 and 245 pretzels.

Step-by-step explanation:

This explanation uses a z-score table where each $z$ entry has two decimal places.

Let $\mu$ represent the mean of a normal distribution of variable $X$ . Let $\sigma$ be the standard deviation of the distribution. The z-score for the observation $x$ would be:

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question,

$\mu = 240$ .
$\sigma = 9.3$ .

Calculate the z-score for $x_1 = 225$ and $x_2 = 245$ . Keep in mind that each entry in the z-score table here has two decimal places. Hence, round the results below so that each contains at least two decimal places.

$\begin{aligned} z_1 &= \frac{x_1 - \mu}{\sigma} \\ &= \frac{225 - 240}{9.3} \approx -1.61\end{aligned}$ .

$\begin{aligned} z_2 &= \frac{x_2 - \mu}{\sigma} \\ &= \frac{245 - 240}{9.3} \approx 0.54\end{aligned}$ .

The question is asking for the probability $P(225 \le X \le 245)$ (where $X$ is between two values.) In this case, that's the same as $P(-1.61 \le Z \le 0.54)$ .

Keep in mind that the probabilities on many z-table correspond to probability of $P(Z \le z)$ (where $Z$ is no greater than one value.) Therefore, apply the identity $P(z_1 \le Z \le z_2) = P(Z \le z_2) - P(Z \le z_1)$ to rewrite $P(-1.61 \le Z \le 0.54)$ as the difference between two probabilities:

$P(-1.61 \le Z \le 0.54) = P(Z \le 0.54) - P(Z \le -1.61)$ .

Look up the z-table for $P(Z \le 0.54)$ and $P(Z \le -1.61)$ :

$P(Z \le 0.54)\approx 0.70540$ .
$P(Z \le -1.61) \approx 0.05370$ .

$\begin{aligned}& P(225 \le X \le 245) \\ &= P\left(\frac{225 - 240}{9.3} \le Z \le \frac{245 - 240}{9.3}\right)\\&\approx P(-1.61 \le Z \le 0.54) \\ &= P(Z \le 0.54) - P(Z \le -1.61)\\ &\approx 0.70540 - 0.05370 \\& \approx 0.65.2 \\ &= 65.2\% \end{aligned}$ .

3 0

3 years ago