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3 years ago

What is the probability that, by guessing at random one can correctly guess the month of birth of a stranger?

Mathematics

2 answers:

diamong [38]3 years ago

7 0

There's 12 months.
Person guesses 1 month.

So, the chances of correctly guessing is 1/12

joja [24]3 years ago

6 0

There is a total of 12 months.....and 1 stranger
so the probability of guessing the birth month of a stranger is 1/12

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Please Help ;-;

zubka84 [21]

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (-8, -8) and (-7, 9). Substitute to the formula of a slope:

$m=\dfrac{9-(-8)}{-7-(-8)}=\dfrac{17}{1}=17$

Put the coordinates of the point (-7, 9) and the value of slope to the point-slope formula:

$y-9=17(x-(-7))\\\\\boxed{y-9=17(x+7)}$

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2 years ago

Ariel made cupcakes to sell. 2/3 of them were chocolate chip,25% of the reminder was strawberry and the rest were 90 vanilla cup

Anton [14]

Answer:

Let x be the total number of cupcakes.

Chocolate chips =

$\frac{2}{3} x$

Strawberry =

$\frac{1}{4} \times \frac{1}{3} \\ = \frac{1}{12} x$

Vanilla =

$x - \frac{2}{3} x - \frac{1}{12}x \\ = \frac{1}{3} x - \frac{1}{12} x \\ = \frac{1}{4} x$

given

$\frac{1}{4} x = 90$

$x = 90 \div \frac{1}{4} \\ = 90 \times 4 \\ = 360$

a)Number of Chocolate chip Cupcakes =

$\frac{2}{3} \times 360 \\ = 240$

b)Number of Strawberry cupcakes =

$\frac{1}{12} \times 360 \\ = 30$

8 0

3 years ago

A fitness club requires a one -time free of $100 and dies of $20 every month .write an expression for the total cost of membersh

nydimaria [60]

The expression would be 20n + 100.

4 0

3 years ago

Read 2 more answers

Find an equation in standard form of the parabola passing through the points.

lana66690 [7]

192
Next time try to add more points bc i worked so hard to answer this

7 0

2 years ago

In a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access

sweet-ann [11.9K]

Answer:

No, the marketing manager was not correct in his claim.

Step-by-step explanation:

We are given that in a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access in their cars.

Suppose that the marketing manager of a car manufacturer claims that the 46.6% is based only on a sample and that 46.6% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than 0.50.

Let p = population proportion of all adult Americans who want car web access

SO, Null Hypothesis, $H_0$ : p $\geq$ 50% {means that the proportion of all adult Americans who want car web access is more than or equal to 0.50}

Alternate Hypothesis, $H_a$ : p < 50% {means that the proportion of all adult Americans who want car web access is less than 0.50}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who indicated that they were somewhat interested or very interested in having web access in their cars = 46.6%

n = sample of Americans = 1005

So, test statistics = $\frac{0.466-0.50}{\sqrt{\frac{0.466(1-0.466)}{1005} } }$

= -2.161

Since in the question we are not given the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z so we sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the proportion of all adult Americans who want car web access is less than 0.50 which means the marketing manager was not correct in his claim.

3 0

3 years ago