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3 years ago

Can someone please explain how to figure this out!!

1 answer:

6 0

2.9% OF the 500 phones are more than likely defective. In order to figure this out you must do 500*0.029 which would equal 14.5 and rounds off to 15. 15 out of the 500 phones will maybe be defective.

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Pls help with these 3 math questions!!!!!!!!

Mariulka [41]

Answer:

The first question- 2-1.8 centimters, just subtract 49.2 from 51

The second question-2-1.052 meters, just divide

The thirs question-3-115, just divide

7 0

2 years ago

Perform the multiplication. Simplify the answers. 2 √30*( √5+ √6+ √10+ √15)

Delicious77 [7]

The simplified expression of $2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})$ is $10\sqrt{6} + 6\sqrt{20}+ 20\sqrt{3} + 30\sqrt{2}$

The expression is given as:

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})$

Expand the expression

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2\sqrt{30} \times \sqrt5+ 2\sqrt{30} \times \sqrt6+ 2\sqrt{30} \times \sqrt{10} + 2\sqrt{30} \times \sqrt{15}$

Factor out 2

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(\sqrt{30} \times \sqrt5+ \sqrt{30} \times \sqrt6+ \sqrt{30} \times \sqrt{10} + \sqrt{30} \times \sqrt{15})$

Combine the radicals

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(\sqrt{150} + \sqrt{180}+ \sqrt{300} + \sqrt{450})$

Expand the expression

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(\sqrt{25 \times 6} + \sqrt{9 \times 20}+ \sqrt{100 \times 3} + \sqrt{225\times 2})$

Evaluate the roots

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(5\sqrt{6} + 3\sqrt{20}+ 10\sqrt{3} + 15 \sqrt{2})$

Expand

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) =10\sqrt{6} + 6\sqrt{20}+ 20\sqrt{3} + 30\sqrt{2}$

Hence, the simplified expression of $2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})$ is $10\sqrt{6} + 6\sqrt{20}+ 20\sqrt{3} + 30\sqrt{2}$

Read more about simplified expressions at:

brainly.com/question/8008182

3 0

2 years ago

In brianna's homeroom class, 9% of the students were born in march and 40% of the students have a blood type of o+.

Firlakuza [10]

Answer:

3.6%

Step-by-step explanation:

Because the two events are independent, their probabilities are multiplied, so the probability of a student being born in March and having a blood type of O+ is 9% * 40% = 0.09 * 0.40 = 0.036 = 3.6%

3 0

1 year ago

Help !!! 8th grade math

Ipatiy [6.2K]

#2=4
#3=16
#4=64
#5=256

4 0

3 years ago

Read 2 more answers

A county is considering rasing the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 m

DochEvi [55]

Answer:

No, at alpha equals 0.10, we do not have enough evidence to support the county's claim.

Step-by-step explanation:

We are given that a county is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 30 miles per hour.

A random sample of 15 vehicles has a mean speed of 31 miles per hour and a standard deviation of 4.7 miles per hour.

Let $\mu$ = true mean speed of the vehicles.

SO, Null Hypothesis, $H_0$ : $\mu \leq$ 30 miles per hour {means that the mean speed of vehicles is lesser than or equal to 30 miles per hour}

Alternate Hypothesis, $H_A$ : $\mu$ > 30 miles per hour {means that the mean speed of vehicles is greater than 30 miles per hour}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean speed of 15 vehicles = 31 mph

s = sample standard deviation = 4.7 mph

n = sample of vehicles = 15

So, test statistics = $\frac{31-30}{\frac{4.7}{\sqrt{15} } }$ ~ $t_1_4$

= 0.824

Hence, the value of test statistics is 0.824.

Now at 0.10 significance level, the t table gives critical value of 1.345 at 14 degree of freedom for right-tailed test. Since our test statistics is less than the critical value of t as 0.824 < 1.345, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean speed of vehicles is lesser than or equal to 30 miles per hour which means that the county's claim is not supported.

4 0

3 years ago