All categories

3 years ago

How many triangles can be constructed with sides measuring 6 cm, 2 cm, and 7 cm?

Mathematics

2 answers:

Katen [24]3 years ago

8 0

One i think i’m not sure tho

Vsevolod [243]3 years ago

4 0

One is the correct answer

You might be interested in

Can anyone explain this to me please

timurjin [86]

Hello there,

To start it off, 2 x 39= 78
So a is correct that would be 2 x 30+9 = 2x39= 78
So b is correct as well. 2 x 40-1= 2 x 39= 78
c is questionable 2 x 30 +9= 60 + 9 = 69
So d is correct 18 + 60 = 78
So e is correct 100-22 = 78

So 2 x 39 is not equal to (2 x 30) +9

Hope this helped!

5 0

3 years ago

Mark had 3 quarts of milk how many pints of milk did he have

mamaluj [8]

Mark had 6 pints of milk.

7 0

4 years ago

Read 2 more answers

Iron deficiency anemia is an important nutritional health problem in the U.S. A dietary assessment was performed on 51 boys 9-11

Bond [772]

Answer:

a) Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

b) $t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

c) $12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

Step-by-step explanation:

Information given

$\bar X=12.50$ represent the mean for the daily iron intake

$s=4.75$ represent the sample deviation

$n=51$ sample size

$\mu_o =14.44$ represent the reference value

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Part a

We want to test if the mean iron intake among the low-income group is different from that of the general population, the system of hypothesis would be:

Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

Part b

Since we don't know the population deviation the statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got

$t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

Part c

The confidence interval would be given by:

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