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3 years ago

0.17x0.4 helpppppppp

Mathematics

1 answer:

sdas [7]3 years ago

5 0

Answer:

The answer is 0.068.

Step-by-step explanation:

0.17 x 0.4 is 0.068

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Samantha sends her son, Barry, to a preschool center on certain days. The cost of preschool is $45 per day along with a fixed mo

Travka [436]

Answer: Option D

D. $880 = 45d + 70$ ; $18\ days$

Step-by-step explanation:

We know that the cost of preschool is $ 45 per day plus a monthly fee of $ 70.

We also know that a total of $ 880 was paid last month

To write an equation that represents this situation, let us call "d" the number of days that Barry attends school

So the cost was:

$45d + 70 = 880$

Now we solve the equation for the variable d

$45d= 880-70$

$d= \frac{810}{45}$

$d= 18\ days$

Therefore answer is the option D

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3 years ago

Roger had $1200. He bought a phone with 0.4 of the money and a helmet with 0.5 of the remaining money. What is the amount of mon

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Answer:

the remaining money is $1189.22

Step-by-step explanation:

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3 years ago

The following table gives results from two groups of students who took a nonproctored test. Use a 0.01 significance level to tes

Nana76 [90]

Answer:

Original claim is $\mu_{1} =\mu_{2}$

Opposite claim is $\mu_{1} \neq \mu_{2}$

Null and alternative hypotheses:

$H_{0}:\mu_{1} = \mu_{2}$

$H_{1} : \mu_{1} \neq \mu_{2}$

Significance level: 0.01

Test statistic:

We can use TI-84 calculator to find the test statistic and P-value. The steps are as follows:

Press STAT and the scroll right to TESTS

Scroll down to 2-SampTTest... and scroll to stats.

Enter below information.

$\bar{x_{1}}=70.29$

$Sx1=22.09$

$n_{1} = 30$

$\bar{x_{2}}=74.26$

$Sx2=18.15$

$n_{2} = 32$

$\mu_{1} \neq \mu_{2}$

Pooled: Yes

Calculate.

The output is in the attachment.

Therefore, the test statistic is:

$t=-0.78$

P-value: 0.4412

Reject or fail to reject: Fail to reject

Final Conclusion: Since the p-value is greater than the significance level, we, therefore, fail to reject the null hypothesis and conclude that the there is sufficient evidence to support the claim that the samples are from populations with the same mean.