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3 years ago

PLZZ HELLO ILL GIVE BRAINLIST

Mathematics

1 answer:

sp2606 [1]3 years ago

3 0

Answer:

X= 1

Step-by-step explanation:

2, + the 2 her brother gives her is too high, its 4.

which means we go lower.

0 can't be right, because 0 + 2 = 2, not 3.

lets try 1.

1 + 2 = 3.

So that means X = 3.

You might be interested in

A normal population has a mean of 19 and a standard deviation of 5.

dangina [55]

Answer:

a) $Z = 1.2$

b) 38.49% of the population is between 19 and 25.

c) 34.46% of the population is less than 17.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

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

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. If we need to find the probability that the measure is larger than X, it is 1 subtracted by this pvalue.

For this problem, we have that

A normal population has a mean of 19 and a standard deviation of 5, so $\mu = 19, \sigma = 5$ .

(a) Compute the z value associated with 25

This is Z when $X = 25$

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

$Z = \frac{25 - 19}{5}$

$Z = 1.2$

(b) What proportion of the population is between 19 and 25?

This is the pvalue of Z when $X = 25$ subtracted by the pvalue of Z when $X = 19$ .

X = 25 has $Z = 1.2$ , that has a pvalue of 0.8849.

X = 19 has $Z = 0$ , that has a pvalue of 0.5000.

So 0.8849-0.500 = 0.3849 = 38.49% of the population is between 19 and 25.

(c) What proportion of the population is less than 17?

This is the pvalue of Z when $X = 17$

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

$Z = \frac{17 - 19}{5}$

$Z = -0.40$

$Z = -0.40$ has a pvalue of 0.3446.

This means that 34.46% of the population is less than 17.

7 0

3 years ago

Of a group of randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wa

Wittaler [7]

Answer:

We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

Step-by-step explanation:

We are given that in a group of randomly selected adults, 160 identified themselves as executives.

n = 160

Also we are given that 42 of executives preferred trucks.

So the proportion of executives who prefer trucks is given by

p = 42/160

p = 0.2625

We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.

We can use normal distribution for this problem if the following conditions are satisfied.

n×p ≥ 10

160×0.2625 ≥ 10

42 ≥ 10 (satisfied)

n×(1 - p) ≥ 10

160×(1 - 0.2625) ≥ 10

118 ≥ 10 (satisfied)

The required confidence interval is given by

$p \pm z\times \sqrt{\frac{p(1-p)}{n} }$

Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.

Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96

$p \pm z\times \sqrt{\frac{p(1-p)}{n} }$

$0.2625 \pm 1.96\times \sqrt{\frac{0.2625(1-0.2625)}{160} }$