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skad [1K]
4 years ago
13

Mrs. Kwan

Mathematics
1 answer:
9966 [12]4 years ago
3 0

One and a half gallons. (1.5 gallons)

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A rectangle has a height of 7a² and a width of a4 + 5a2 + 4.
harkovskaia [24]

Answer:

( {7a}^{2}  \times  {a}^{4} ) +  ({7a}^{2}  \times  {5a}^{3})  + ( {7a}^{2}   \times 4) =  {7a}^{6}  +  {35a}^{5}  +  {28a}^{2}

8 0
4 years ago
Please help me and it is not 12.<br> Value of
topjm [15]
Would it be 18 if it’s not 12? I’m sorry i don’t know but i’m trying to help as much as possible
3 0
3 years ago
a House sits on a 5/8 acre lot and 1/2 of the lot is lawn. 2/3 of lawn was mowed. how much was mowed?
eimsori [14]

5/8 x 1/2 = 5/16 acre is lawn

5/16 x 2/3 = 10/48 reduces to 5/24 of the lawn was mowed

3 0
3 years ago
Sampling every 100th invoice would be an example of:_______a. Randomization.b. Proxemics.c. Stratification.d. JRP.e. Observation
gregori [183]

Answer: C. Stratification

Step-by-step explanation: There are diverse samping techniques employed for statistical purposes depending on the tipe and nature of data available or other factors. When a data population is sampled in sects such that the entire data is divided into a equals groups at a fixed intervals based on a certain criteria, it is called stratified sampling. Here, data is divided into strata, with each stratum containing a fixed amount of samples with each partition taking place at regular interval. In the scenario stratification process is employed by sampling at a fixed and regular interval, however, the random nature of the data is not violated.

8 0
4 years ago
In the production of a plant, a treatment is being evaluated to germinate seed. From a total of 60 seed it was observed that 37
djyliett [7]

Answer:

More than 50% would germinate

Step-by-step explanation:

Given that in the production of a plant, a treatment is being evaluated to germinate seed. From a total of 60 seed it was observed that 37 of them germinated

Let us check whether more than 50% will germinate using hypothesis test

H_0: p = 0.50\\H_a: p>0.50\\

(right tailed test)

Sample proportion p =\frac{37}{60} =0.617\\q = 0.383\\Std error = \sqrt{\frac{pq}{n} } =0.0628

p difference = 0.117

Test statistic Z = p difference/std error = 1.864

p value =0.0312

Since p value <0.05 our significance level of 5% we reject null hypothesis

It is  possible to claim that most of the seed will germinate (i.e. more than 50%)

7 0
3 years ago
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