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3 years ago

Two sides of a triangle measure 25cm and 35cm. which of the following could be the measure of the 3rd side

Mathematics

1 answer:

Bumek [7]3 years ago

8 0

11 cm? Im not that good at this.. If im correct any of those would work

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Tim's bean sprout grew 3/ inches. Teegan's bean sprout grew 2 3/4 inches. How many more inches did Tim's bean sprout grow than T

ki77a [65]

Answer:

Tim's bean sprout grow by more than $\frac{5}{8} \text{ inches}$ .

Step-by-step explanation:

We are given that Tim's bean sprout grew 3 3/8 inches. Teegan's bean sprout grew 2 3/4 inches.

We have to find how many more inches did Tim's bean sprout grow than Teegan's.

Firstly, converting both the mixed fractions in improper fraction get;

Tim's bean sprout grew = $3\frac{3}{8}$ = $\frac{27}{8} \text{ inches}$

Teegan's bean sprout grew = $2\frac{3}{4}$ = $\frac{11}{4} \text{ inches}$

Since the denominator of both the fractions is not the same, so we can't compare them both as which is larger or smaller.

Tim's bean sprout grew = $\frac{27\times 1}{8\times 1} = \frac{27}{8} \text{ inches}$

Teegan's bean sprout grew = $\frac{11 \times 2}{4\times 2}=\frac{22}{8} \text{ inches}$

Now, we can clearly see that Tim's bean sprout grew more than Teegan's bean sprout.

So, Tim's bean sprout grow by more than = $\frac{27}{8}- \frac{22}{8}$ = $\frac{5}{8} \text{ inches}$

Hence, Tim's bean sprout grow by more than $\frac{5}{8} \text{ inches}$ .

8 0

2 years ago

JPS

RUDIKE [14]

The correct answer is true

3 0

2 years ago

Im confused about this question can someone solve it for me

Brums [2.3K]

Answer:

Step-by-step explanation:

you just foil them together and get that

3 0

2 years ago

Given a normal distribution with mu equals 100μ=100 and sigma equals 10 commaσ=10, complete parts (a) through (d). loading...

m_a_m_a [10]

I believe the parts are:

A. What is the probability that Upper X greater than 95X>95?

B. What is the probability that Upper X less than 75X<75?

C.What is the probability that Upper X less than 85X<85 and Upper X greater than 125X>125?

D. 95% of the values are between what two X-values (symmetrically distributed around the mean)?

<u>Solution:</u>

We use the equation for z score:

z = (X – μ) / σ

Then use the standard normal probability tables to locate for the value of P at indicated z score value.

A. Z = (95 – 100) / 10 = - 0.5

Using the tables, the probability at z = -0.5 using right tailed test is:

P = 0.6915

B. Z = (75 – 100) / 10 = - 2.5

Using the tables, the probability at z = -2.5 using left tailed test is:

P = 0.0062

C. Z = (85 – 100) / 10 = - 1.5

Using the tables, the probability at z = -2.5 using left tailed test is:

P = 0.0668

Z = (125 – 100) / 10 = 2.5

Using the tables, the probability at z = 2.5 using right tailed test is:

P = 0.0062

So the probability that 85<X and X>125 is:

P(total) = 0.0668 + 0.0062

P(total) = 0.073

D. P(left) = 0.025, Z = -1.96

P(right) = 0.975, Z = 1.96

The X’s are calculated using the formula:

X = σz + μ

At Z = -1.96

X = 10 (-1.96) + 100 = 80.4

At Z = 1.96

X = 10 (1.96) + 100 = 119.6

<span>So 95% of the values are between 80.4 and 119.6 (80.4< X <119.6).</span>

7 0

3 years ago

<-- 50 points No Trolls please

vichka [17]

Answer:

the correct answer is 1 because it looks like they are doing it by odd numbers.

hope this helps :)

Step-by-step explanation:

5 0

2 years ago

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