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3 years ago

What is the LCM of 9 and 15?A) 15 B) 30 C) 45 D) 54

Mathematics

2 answers:

frez [133]3 years ago

8 0

Answer:

Step-by-step explanation:

The multiples of 9 are : 9, 18, 27, 36, 45, 54, 63,......

The multiples of 15 are : 15, 30, 45, 60, 75, ........

The LCM of 9 and 15 is 45

melamori03 [73]3 years ago

6 0

C is the most reasonable answer

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a sports team donates 157 tickets to a sporting event,with tickets to be shared equally among 9 classrooms.How many tickets does

Valentin [98]

Answer:

Each class should get 17 tickets. 4 tickets will be left over.

Step-by-step explanation:

157/9=17.444... -> 17 per class

17 x 9 = 153

157-153= 4 left

5 0

3 years ago

Help:<br> <img src="https://tex.z-dn.net/?f=1.%5C%20%26x%5E4-61x%5E2%2B900%3D0%20%5C%5C%202.%20%5C%20%26%20x%5E4-25x%5E2%2B144%3

yawa3891 [41]

$\bf{1) \ x^4-61x^2+900=0 }$

Applying the factorization method, for example:

$\bf{ 0=(x^2)^2-61(x^2)+900=(x^2-36)(x^2-25)}$

$\bf{\\ &=(x+6)(x-6)(x+5)(x-5); }$

$\bf{ x_1=6 \quad x_2=-6 \quad \ \ \ x_3=5 \quad x_4=5 }$

$\bf{2) \ x^4-25x^2+144=0 }$

Applying the factorization method:

$\bf{0&=(x^2)^2-25(x^2)+144=(x^2-16)(x^2-9) }$

$\bf{ =(x-4)(x+4)(x-3)(x+3); }$

$\bf{ x_1=4 \quad x_2=-4 \quad x_3=3 \quad x_4=-3 }$

7 0

2 years ago

in 1991, a gallup poll reported this percent to be 79%. using the data from this poll, test the claim that the percent of driver

snow_lady [41]

We reject our null hypothesis, H₀, at a level of significance of =0.01 since the P-value is less than that threshold. There is compelling statistical data to indicate that since 1991, the proportion of drivers who love driving has decreased.

Given,

The Pew Research Center recently polled n=1048 U.S. drivers and found that 69% enjoyed driving their automobiles.

In 1991, a Gallup poll reported this percentage to be 79%. using the data from this poll, test the claim that the percentage of drivers who enjoy driving their cars has declined since 1991.

To report the large-sample z statistic and its p-value,

Null hypothesis,

H₀ : p = 0.79

Alternative hypothesis,

Ha : p < 0.79

Level of significance, α = 0.01

Under H₀

Test statistic,

$Z_0= \frac{\hat p - p}{\sqrt{\frac{p(1-p)}{n} } }$

Z₀ = -7.948

The alternative hypothesis(Ha) is left-tailed, so the P-value of the test is given by

P-value = P(z <-7.945)

= 0.000 (from z-table)

Since the P-value is smaller than given level of significance, α=0.01 we reject our null hypothesis, H₀, at α=0.0.1 level Strong statistical evidence to conclude that the percentage of drivers who enjoy driving their cars has declined since 1991.

To learn more about hypothesis click here:

brainly.com/question/17173491

#SPJ4

5 0

1 year ago

if light bulbs are 4 for $2.00 , how much will 12 light bulbs cost , round to the nearest cent if necessary?

Rufina [12.5K]

If 4 are $2.00 , then x3 to get 12 light bulbs and 2x3=6 so it's $6.00

8 0

3 years ago

If Upper X overbar equals 62, Upper S equals 8, and n equals 36, and assuming that the population is normally distributed, c

marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=62$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=36-1=35$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that $t_{\alpha/2}=2.72$

Now we have everything in order to replace into formula (1):

$62-2.72\frac{8}{\sqrt{36}}=58.373$

$62+2.72\frac{8}{\sqrt{36}}=65.627$

So on this case the 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

3 0

3 years ago