All categories

2 years ago

Evaluate: (7.88)(-1.5)

Mathematics

2 answers:

Svetach [21]2 years ago

7 0

(7.88)(-1.5) = 7.88•-1.5
You would multiply and get
-11.82

Misha Larkins [42]2 years ago

6 0

Answer:

7.88 by − 1.5 = − 11.82

Step-by-step explanation:

Multiply the two

You might be interested in

Multiply. Write your answer in simplest form.

Mariana [72]

Answer:

The answer in lowest terms is 15/17

Step-by-step explanation:

Hope that helps!

3 0

2 years ago

Riding an elevator down 3 floors. Is that positive or negative

Artyom0805 [142]

That is negative since the elevator is decreasing. If the elevator was going up, it would be positive.

6 0

3 years ago

In the 1980’s, a clinical trial was conducted to determine if taking an aspirin daily reduced the incidence of heart attacks. Of

zubka84 [21]

There is enough evidence to conclude that taking aspirin cannot reduces the chance of cancer.

Given sample size of patients take aspirin 11037, sample size of patients who have assigned placebo group be 11034. 104 doctors who take aspirin had a heart attack, 189 doctors had placebo had heart attacks.

First we have to form hypothesis.

$H_{0} :p{1} -p_{2} =0$

$H_{1}:p_{1} -p_{2} < 0$

We have to find the respective probabilities.

$p_{1}$ =104/11037

=0.0094

$p_{2}$ =189/11034

=0.0171

Now their respective margin of errors.

$s_{1}$ = $\sqrt{ {(0.0094*0.9906)/11037}$

=0.0009

$s_{2}$ = $\sqrt{0.0171*0.9829}$

=0.0011

Hence the distribution of the differences,they are given by:

p= $p_{1} -p_{2}$

=0.0094-0.0171

=-0.0077

S= $\sqrt{s_{1} ^{2}+s_{2} ^{2} }$

= $\sqrt{(0.0009)^{2} +(0.0011)^{2} }$

=0.00305

z=(p -f)/S (In which f=0 is the value tested at the null hypothesis)

=(-0.0077-0)/0.00305

=-2.52

p value will be 0.005.

p value of 0.05 significance level.

z=1.96.

1.96>0.005

So we will reject the null hypothesis which means it cannot reduce the whole chance of becomming a heart attack.

Hence there is enough evidence to conclude that taking aspirin cannot reduces the chance of cancer.

Learn more about t test at brainly.com/question/6589776

#SPJ4

6 0

2 years ago

A poll found that 53% of a sample of 1170 teens in a certain large country go online several times a day. (Treat this sample as

posledela

Answer:(a) margin error = 2.4%

(b) The margin error gives the measure in percentage of how the population parameter determined differ from the real population statistics or value.

(c) in 90% of the samples of teens in the country, the percent who go online several times a day will be within 50.6% and 55.4%. of the estimated 100%

Step-by-step explanation:

Using the proportion formulae

Margin error = z √p(1-p)/n

n= 1170, p = 53% = 0.53, 1-p = 0.47

and the z value at 90% C.I = 1.645

M error= 1.645 √0.53×0.47/1170

Margin error = 0.024 = 0.024 ×100

Margin error = 2.4%

53 - 2.4 = 50.6% and 53 + 2.4 = 55.4%

In other words 90% of the time: the number of teens who go online several time a day will be between 50.6 and 55.4%.

8 0

3 years ago

Find the additive inverse. <br> 6

Irina18 [472]

-6 it is just the opposite of the number. So a positive would be a negative and a negative a positive. So the answer to your question is -6.

6 0

3 years ago

Evaluate: (7.88)(-1.5)​

Evaluate: (7.88)(-1.5)