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3 years ago

Ben adopted 2 cats and 3 dogs. He spent a total of $455. If each dog cost him $82, how much did one cat cost him?

Mathematics

1 answer:

monitta3 years ago

5 0

Answer:$104.5

Step-by-step explanation:

cats=c

Dogs=d

2c+3d=455

d=82

Substitute d=82 in 2c+3d=455

2c+3(82)=455

2c+246=455

Collect like terms

2c=455-246

2c=209

Divide both sides by 2

2c/2=209/2

c=104.5

1 cat cost $104.5

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2 years ago

Suppose that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage

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Answer:

We conclude that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

We are given that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%.

Let p = population percentage of blue candies

So, Null Hypothesis, $H_0$ : p = 29% {means that the percentage of blue candies is equal to 29%}

Alternate Hypothesis, $H_A$ : p $\neq$ 29% {means that the percentage of blue candies is different from 29%}

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

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

where, $\hat p$ = sample proportion of blue coloured candies = 28%

n = sample of colored candies = 100

So, the test statistics = $\frac{0.28-0.29}{\sqrt{\frac{0.29(1-0.29)}{100} } }$

= -0.22

The value of the z-test statistics is -0.22.

Also, the P-value of the test statistics is given by;

P-value = P(Z < -0.22) = 1 - P(Z $\leq$ 0.22)

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Now, at a 0.10 level of significance, the z table gives a critical value of -1.645 and 1.645 for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of z, so we insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the percentage of blue candies is equal to 29%.

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