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Gelneren [198K]
3 years ago
5

Could someone help me with math please and please txt back

Mathematics
1 answer:
Pani-rosa [81]3 years ago
3 0
What do you need help on?

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If two friends invested in a ratio of 2:3 and earned a profit of Rs. 25,000 on their investments, then the smaller share of the
podryga [215]
Sorry I just needed the points
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3 years ago
Shobo's mother's present age is six times Shobo's present age. Shobo's age five
il63 [147K]

Answer:

Step-by-step explanation:

Shobo's age = x years

His mother's age = 6 times of

Shobo's age

= 6x years

Five years from now

________________

Shobo's age = ( x + 5 ) years

According to the problem given,

x + 5 = one by third of his mother's

present age

x + 5 = 6x / 3

x + 5 = 2x

5 = 2x - x

5 = x

Therefore ,

Shobo's present age = x = 5 years

His mother's age = 6x

= 6 × 5

= 30 years

4 0
3 years ago
Identify the slope (m) and y intercept for -0.6x+y=-4​
Natalija [7]

Answer:

Slope is -0.6

y-intercept is -4

Step-by-step explanation:

y=mx+b

8 0
3 years ago
Read 2 more answers
A bunch of friends went to the snack shake for lunch. The first family ordered 4 hamburger and 4 orders of fries for 9$. The sec
nexus9112 [7]

Answer:

$1.50 for hamburgers, and $0.75 for fries.


Hope this helps you! Have an amazing day :)                                                                                  

8 0
3 years ago
At a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective pa
11Alexandr11 [23.1K]

Answer:

Probability that fewer than 2 of these parts are defective is 0.604.

Step-by-step explanation:

We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.

A random sample of 7 parts produced by this machine is chosen.

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 7 parts

            r = number of success = fewer than 2

           p = probability of success which in our question is % of defective

                 parts produced by one of the machine, i.e; 19%

<em>LET X = Number of parts that are defective</em>

<u>So, it means X ~ Binom(n = 7, p = 0.19)</u>

Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)

    P(X < 2) = P(X = 0) + P(X = 1)

                  =  \binom{7}{0}\times 0.19^{0} \times (1-0.19)^{7-0}+ \binom{7}{1}\times 0.19^{1} \times (1-0.19)^{7-1}

                  =  1 \times 1 \times 0.81^{7} +7 \times 01.9^{1} \times 0.81^{6}

                  =  <u>0.604</u>

<em>Therefore, the probability that fewer than 2 of these parts are defective is 0.604.</em>

8 0
2 years ago
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