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Lyrx [107]
2 years ago
5

Help please asap tysm! 10 brainly points! Will mark brainliest if you put in a random answer you will be reported! Tysm! <3 N

eed this asap please tysm!

Mathematics
1 answer:
kkurt [141]2 years ago
8 0

Answer:

B

Step-by-step explanation:

Dividing by a number is the same as multiplying by the reciprocal (one over that number)

so z/144 is equal to z * 1/144

You might be interested in
2x+15=145<br> Solve for x
Ganezh [65]

Answer:

Step-by-step explanation:2x=145-15

2x=130

X=130/2

X=65

6 0
3 years ago
One batch of mooncakes contains 1/4 tablespoons of kansui. How many tablespoons of kansui are
Ghella [55]

From the division property, the required kansui is 5/32 tablespoon.

What is division?

One of the four fundamental arithmetic operations, or the process by which two or more numbers are added together to create a new number, is division. Multiplication, addition, and subtraction round out the list of operations.

Given that a batch of mooncake contains 1/4 tablespoon of kansui.

Given batch of mooncake is 1 3/5 = 8/5

Therefore,

\frac{\frac{1}{4}}{\frac{8}{5}}\\=\frac{1}{4} \times \frac{5}{8}\\=\frac{5}{32}

Hence, the required kansui is 5/32 tablespoon.

To learn more about division from the given link

brainly.com/question/25289437

#SPJ9

3 0
1 year ago
I need this answer, please?
Anvisha [2.4K]

Answer:

4.23

Step-by-step explanation:

3.68 + 0.21 + 0.25 + 0.01 + 0.08 = 4.23

4 0
3 years ago
The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp
andreyandreev [35.5K]

Answer:

(a) X ~ N(\mu=63, \sigma^{2} = 13^{2}).

    \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = <u><em>amount of syrup that people put on their pancakes</em></u>

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

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

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

So, the distribution of X ~ N(\mu=63, \sigma^{2} = 13^{2}).

Let \bar X = <u><em>sample mean amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the sample mean is given by;

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

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

            n = sample of people = 43

So, the distribution of \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

   P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X \leq 61.4 mL)

  P(X < 62.8 mL) = P( \frac{X-\mu}{\sigma} < \frac{62.8-63}{13} ) = P(Z < -0.02) = 1 - P(Z \leq 0.02)

                                                           = 1 - 0.50798 = 0.49202

  P(X \leq 61.4 mL) = P( \frac{X-\mu}{\sigma} \leq \frac{61.4-63}{13} ) = P(Z \leq -0.12) = 1 - P(Z < 0.12)

                                                           = 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < \bar X < 62.8 mL)

   P(61.4 mL < \bar X < 62.8 mL) = P(\bar X < 62.8 mL) - P(\bar X \leq 61.4 mL)

  P(\bar X < 62.8 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{62.8-63}{\frac{13}{\sqrt{43} } } ) = P(Z < -0.10) = 1 - P(Z \leq 0.10)

                                                           = 1 - 0.53983 = 0.46017

  P(\bar X \leq 61.4 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{61.4-63}{\frac{13}{\sqrt{43} } } ) = P(Z \leq -0.81) = 1 - P(Z < 0.81)

                                                           = 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

4 0
3 years ago
PLEASEE HELP!!<br><br> Write and equation for this graph in slope- intercept form.
ella [17]
-2/1. You find two point that go on the line and if it’s going down it’s negative
8 0
2 years ago
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