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strojnjashka [21]

2 years ago

6

A store selling school supplies advertises a bundle deal. The consumer can pick a backpack, a binder, a pack of pencils, and not

ebook paper for a set price. There are 5 types of backpacks, 5 types of binders, 3 types of pencils, and 2 types of notebook paper. How many outcomes are possible in this bundle?

2 answers:

LiRa [457]2 years ago

6 0

5*3*2=30 combinations r possible

aniked [119]2 years ago

5 0

Answer:

150

Step-by-step explanation:

Given :There are 5 types of backpacks, 5 types of binders, 3 types of pencils, and 2 types of notebook paper.

To Find: How many outcomes are possible in this bundle?

Solution:

No. of backpack= 5

No. of binders = 5

No. of pencils = 3

No. of type of notebook paper = 2

Since we are given that The consumer can pick a backpack, a binder, a pack of pencils, and notebook paper for a set price.

So, No. of possible outcomes in this bundle = $5 \times 5\times 3 \times 2$

= $150$

Hence there are 150 possible outcomes.

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Bonita is testing prototype model rocket engines. To be considered a successful launch, the rocket must reach a height of at lea

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Answer:

(0, 33) : Yes

(4.8, 30.5) : Yes

Step-by-step explanation:

height of at least 24 ft : y >= 24

horizontal distance of no more than 10 ft. : 0 <= x <= 10

The vertical height must be at least three times as high as the horizontal distance. No rocket should go higher than 33 ft.:

y >= 3x and y <= 33

So Constraints:

24 <= y <= 33

y >= 3x

0 <= x <= 10

Now check to see if those coordinate points meet all the constraints

(0, 33) : Yes

(4.8, 30.5) : Yes

(9, 26) : No (because 26 is less than (3 * 9) = 27)

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(2, 22): No (because 22 <24, must be 24 <= y <= 33)

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2 years ago

The random variable X is normally distributed with mean 82 and standard deviation 7.4. Find the value of q such that P(82 − q &l

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P(82 - q < x < 82 + q) = 0.44
P(x < 82 + q) - P(82 - q) = 0.44
P(z < (82 + q - 82)/7.4 - P(z < (82 - q - 82)/7.4) = 0.44
P(z < q/7.4) - P(z < -q/7.4) = 0.44
P(z < q/7.4) - (1 - P(z < q/7.4) = 0.44
P(z < q/7.4) - 1 + P(z < q/7.4) = 0.44
2P(z < q/7.4) - 1 = 0.44
2P(z < q/7.4) = 1.44
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1 year ago

4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

pochemuha

Answer:

No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

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The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

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Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

So

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

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2 years ago

Read 2 more answers